€¦ · ordinary differential equations definitions classification basic examples definitions...

Ordinary differential equations

Computational Neuroscience. Session 1-2

Dr. Marco A Roque Sol

05/29/2018

Dr. Marco A Roque Sol Computational Neuroscience. Session 1-2

Ordinary differential equationsDefinitionsClassificationBasic Examples

Definitions

Differential Equations

A differential equation is any equation which containsderivatives, either ordinary or partial derivatives of an unknownfunction.

Ordinary and Partial Differential EquationsA differential equation is called an ordinary differential equation,abbreviated by ODE, if it has only ordinary derivatives in it,having the form:

F (y (n), y (n−1), ..., y ′, y(t), t) = 0

Likewise, a differential equation is called a partial differentialequation, abbreviated by PDE, if it has partial derivatives in it.



Definitions


A differential equation

is any equation which containsderivatives, either ordinary or partial derivatives of an unknownfunction.


F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions


A differential equation is any equation which containsderivatives,

either ordinary or partial derivatives of an unknownfunction.


F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions



Ordinary and Partial Differential Equations

A differential equation is called an ordinary differential equation,abbreviated by ODE, if it has only ordinary derivatives in it,having the form:

F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions



Ordinary and Partial Differential EquationsA differential equation

is called an ordinary differential equation,abbreviated by ODE, if it has only ordinary derivatives in it,having the form:

F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions



Ordinary and Partial Differential EquationsA differential equation is called an ordinary differential equation,abbreviated by ODE,

if it has only ordinary derivatives in it,having the form:

F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0

Likewise, a differential equation

is called a partial differentialequation, abbreviated by PDE, if it has partial derivatives in it.



Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0

Likewise, a differential equation is called a partial differentialequation,

abbreviated by PDE, if it has partial derivatives in it.



Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0

Likewise, a differential equation is called a partial differentialequation, abbreviated by PDE,

if it has partial derivatives in it.



Definitions




F (y (n), y (n−1), ..., y ′, y(t), t) = 0




Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics,

if an object of mass m is moving with acceleration aand being acted on with force F then NewtonâAZs Second Lawtells us.

F = ma

To see that this is in fact a differential equation. First, rememberthat we can rewrite the acceleration, a, in one of two ways.

a =dvdt

or a =d2udt2

Where v is the velocity of the object and u is the positionfunction of the object at any time t . We should also rememberat this point that the force, F may also be a function of time,velocity, and/or position.



Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics, if an object of mass m

is moving with acceleration aand being acted on with force F then NewtonâAZs Second Lawtells us.

F = ma


a =dvdt

or a =d2udt2




Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics, if an object of mass m is moving with acceleration a

and being acted on with force F then NewtonâAZs Second Lawtells us.

F = ma


a =dvdt

or a =d2udt2




Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics, if an object of mass m is moving with acceleration aand being acted on with force F

then NewtonâAZs Second Lawtells us.

F = ma


a =dvdt

or a =d2udt2




Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics, if an object of mass m is moving with acceleration aand being acted on with force F then NewtonâAZs Second Lawtells us.

F = ma


a =dvdt

or a =d2udt2




Definitions


F = ma

To see that this is in fact a differential equation.

First, rememberthat we can rewrite the acceleration, a, in one of two ways.

a =dvdt

or a =d2udt2




Definitions


F = ma

To see that this is in fact a differential equation. First, rememberthat we can rewrite the acceleration, a,

in one of two ways.

a =dvdt

or a =d2udt2




Definitions


F = ma


a =dvdt

or a =d2udt2




Definitions


F = ma


a =dvdt

or

a =d2udt2




Definitions


F = ma


a =dvdt

or a =d2udt2




Definitions


F = ma


a =dvdt

or a =d2udt2

Where v is the velocity of the object

and u is the positionfunction of the object at any time t . We should also rememberat this point that the force, F may also be a function of time,velocity, and/or position.



Definitions


F = ma


a =dvdt

or a =d2udt2

Where v is the velocity of the object and u is the positionfunction of the object at any time t .

We should also rememberat this point that the force, F may also be a function of time,velocity, and/or position.



Definitions


F = ma


a =dvdt

or a =d2udt2

Where v is the velocity of the object and u is the positionfunction of the object at any time t . We should also rememberat this point that the force,

F may also be a function of time,velocity, and/or position.



Definitions


F = ma


a =dvdt

or a =d2udt2

Where v is the velocity of the object and u is the positionfunction of the object at any time t . We should also rememberat this point that the force, F may also be a function of time,velocity, and/or position.Dr. Marco A Roque Sol Computational Neuroscience. Session 1-2


Definitions

mdvdt

= F (t , v) or md2udt2 = F (t , u, v)

More examples of differential equations

2y ′′ + 3y ′ − 5y = 0

cos(y)d2ydx2 − (1 + y)

dydx

+ y3e−y = 0

y (4) + 5y ′′′ − 4y ′′ + y = sin(x)

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2

∂u3

∂2x∂t= 1 +

∂u∂y



Classification

Order

The order of a differential equation is the largest derivativepresent in the differential equation. The equation

mdvdt

= F (t , v)

is a first order differential equation, the equations

md2udt2 = F (t , u, v)

2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0



Classification

OrderThe order

of a differential equation is the largest derivativepresent in the differential equation. The equation

mdvdt

= F (t , v)



2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0



Classification

OrderThe order of a differential equation

is the largest derivativepresent in the differential equation. The equation

mdvdt

= F (t , v)



2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0



Classification

OrderThe order of a differential equation is the largest derivativepresent in the differential equation.

The equation

mdvdt

= F (t , v)



2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0



Classification

OrderThe order of a differential equation is the largest derivativepresent in the differential equation. The equation

mdvdt

= F (t , v)



2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0



Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2

are second order differential equations, the equation∂u3

∂2x∂t= 1 +

∂u∂y

is a third order differential equation and finally, the equation

y (4) + 5y ′′′ − 4y ′′ + y = sin(x)

is a fourth order differential equation.

Note that the order does not depend on whether or not you’vegot ordinary or partial derivatives in the differential equation.



Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2

are second order differential equations, the equation

∂u3

∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)





Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2


∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)





Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2


∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)


Note that the order

does not depend on whether or not you’vegot ordinary or partial derivatives in the differential equation.



Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2


∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)


Note that the order does not depend on whether or not

you’vegot ordinary or partial derivatives in the differential equation.



Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2


∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)


Note that the order does not depend on whether or not you’vegot ordinary or

partial derivatives in the differential equation.



Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2


∂2x∂t= 1 +

∂u∂y


y (4) + 5y ′′′ − 4y ′′ + y = sin(x)





Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β

is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t)

which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval.

It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that

the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals

andthese intervals can impart some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart

some important information aboutthe solution.



Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.



Classification

Linear Differential Equations

A linear differential equation is any differential equation thatcan be written in the following form.

an(t)y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = g(t)

The important thing to note about linear differential equations isthat there are no products of the function and its derivatives,and neither the function or its derivatives occur to any powerother than the first power.



Classification


A linear differential equation

is any differential equation thatcan be written in the following form.





Classification


A linear differential equation is any differential equation thatcan be written

in the following form.





Classification







Classification




The important thing to note

about linear differential equations isthat there are no products of the function and its derivatives,and neither the function or its derivatives occur to any powerother than the first power.



Classification




The important thing to note about linear differential equations

isthat there are no products of the function and its derivatives,and neither the function or its derivatives occur to any powerother than the first power.



Classification




The important thing to note about linear differential equations isthat there are no products

of the function and its derivatives,and neither the function or its derivatives occur to any powerother than the first power.



Classification




The important thing to note about linear differential equations isthat there are no products of the function and its derivatives,and

neither the function or its derivatives occur to any powerother than the first power.



Classification




The important thing to note about linear differential equations isthat there are no products of the function and its derivatives,and neither the function or its derivatives

occur to any powerother than the first power.



Classification




The important thing to note about linear differential equations isthat there are no products of the function and its derivatives,and neither the function or its derivatives occur to any power

other than the first power.



Classification







Classification

The coefficients an(t), ..., a0(t), g(t)

can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0

is a non-linear equation. Note that we can’t classify Newton’sSecond Law equations since we do not know what form thefunction F has. These could be either linear or non-lineardepending on F .



Classification

The coefficients an(t), ..., a0(t), g(t) can be any function.

Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and

its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining

if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear.

If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation

cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called

a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation.

In the examples above only


dydx

+ y3e−y = 0




Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0




Classification



dydx

+ y3e−y = 0

is a non-linear equation.

Note that we can’t classify Newton’sSecond Law equations since we do not know what form thefunction F has. These could be either linear or non-lineardepending on F .



Classification



dydx

+ y3e−y = 0

is a non-linear equation. Note that we can’t classify Newton’sSecond Law equations

since we do not know what form thefunction F has. These could be either linear or non-lineardepending on F .



Classification



dydx

+ y3e−y = 0

is a non-linear equation. Note that we can’t classify Newton’sSecond Law equations since we do not know

what form thefunction F has. These could be either linear or non-lineardepending on F .



Classification



dydx

+ y3e−y = 0

is a non-linear equation. Note that we can’t classify Newton’sSecond Law equations since we do not know what form thefunction F has.

These could be either linear or non-lineardepending on F .



Classification



dydx

+ y3e−y = 0




Basic Examples

Differential Equations in Physics

Schrodinger’s Equation.

− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t

Is the starting point for non-relativistic Quantum Mechanics,here the solution Ψ(r , t) is the wave function.

In the one-dimensional case, If we substitute into the aboveequation, the proposed solution

Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t

Is the starting point for non-relativistic Quantum Mechanics,

here the solution Ψ(r , t) is the wave function.


Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t

Is the starting point for non-relativistic Quantum Mechanics,here the solution Ψ(r , t)

is the wave function.


Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t



Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t


In the one-dimensional case,

If we substitute into the aboveequation, the proposed solution

Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t


In the one-dimensional case, If we substitute into the aboveequation,

the proposed solution

Ψ(x , t) = ψ(x)φ(t)



Basic Examples



− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t



Ψ(x , t) = ψ(x)φ(t)



Basic Examples

we get

− h2

2mψ′′(x)φ(t) + V (x)ψ(x)φ(t) = i hψ(x)φ′(t)

Dividing the whole equation by ψ(x)φ(t) we get

− h2

2mψ′′(x)ψ(x)

+ V (x) = i hφ′(t)φ(t)

= constant = E

So we obtain two ODE’s.



Basic Examples

we get

− h2



− h2

2mψ′′(x)ψ(x)

+ V (x) =

i hφ′(t)φ(t)

= constant = E




Basic Examples

we get

− h2



− h2

2mψ′′(x)ψ(x)

+ V (x) = i hφ′(t)φ(t)

=

constant = E




Basic Examples

we get

− h2



− h2

2mψ′′(x)ψ(x)

+ V (x) = i hφ′(t)φ(t)

= constant =

E




Basic Examples

we get

− h2



− h2

2mψ′′(x)ψ(x)

+ V (x) = i hφ′(t)φ(t)

= constant = E




Basic Examples

First,

The Time Independent Schrodinger Equation

− h2

2md2ψ(x)

dt2 + V (x)ψ(x) = Eψ(x)

and second, The Energy Eigenvalue Equation

i hdφ(t)

dt= Eφ(t)



Basic Examples

First, The Time Independent Schrodinger Equation

− h2

2md2ψ(x)

dt2 + V (x)ψ(x) = Eψ(x)


i hdφ(t)

dt= Eφ(t)



Basic Examples


− h2

2md2ψ(x)

dt2 + V (x)ψ(x) = Eψ(x)

and second,

The Energy Eigenvalue Equation

i hdφ(t)

dt= Eφ(t)



Basic Examples


− h2

2md2ψ(x)

dt2 + V (x)ψ(x) = Eψ(x)


i hdφ(t)

dt= Eφ(t)



Basic Examples

Maxwell’s equations:

∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations for ClassicalElectromagnetism, here the solutions E and B represent theelectrical and magnetic fields.



Basic Examples


∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set

of fundamental equations for ClassicalElectromagnetism, here the solutions E and B represent theelectrical and magnetic fields.



Basic Examples


∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations

for ClassicalElectromagnetism, here the solutions E and B represent theelectrical and magnetic fields.



Basic Examples


∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations for ClassicalElectromagnetism,

here the solutions E and B represent theelectrical and magnetic fields.



Basic Examples


∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations for ClassicalElectromagnetism, here the solutions E and B

represent theelectrical and magnetic fields.



Basic Examples


∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations for ClassicalElectromagnetism, here the solutions E and B represent theelectrical and magnetic fields.



Basic Examples

In the vacuum

ρ = J = 0. Maxwell Equations can be written as

∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2

Each one of these is 3 PDE’s. Consider the one-dimensionalcase for the Electrical Field

∂2E∂x2 = µ0ε0

∂2E∂t2

Now, suppose that E(x , t) = X(x)T (t) making this substitutionwe get



Basic Examples

In the vacuum ρ = J = 0. Maxwell Equations can be written as

∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2


∂2E∂x2 = µ0ε0

∂2E∂t2




Basic Examples


∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2

Each one of these is 3 PDE’s.

Consider the one-dimensionalcase for the Electrical Field

∂2E∂x2 = µ0ε0

∂2E∂t2




Basic Examples


∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2


∂2E∂x2 = µ0ε0

∂2E∂t2




Basic Examples


∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2


∂2E∂x2 = µ0ε0

∂2E∂t2

Now, suppose that E(x , t) = X(x)T (t)

making this substitutionwe get



Basic Examples


∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2


∂2E∂x2 = µ0ε0

∂2E∂t2




Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)

dividing by X(x)T (t)

X′′(x)X(x)

= µ0ε0T ′′(t)T (t)

= constant = a

Thus, we have to solve a couple of ODE’s

d2X(x)dx2 = aX(x); µ0ε0T ′′(t) = aT (t)



Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)


X′′(x)X(x)

=

µ0ε0T ′′(t)T (t)

= constant = a





Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)


X′′(x)X(x)

= µ0ε0T ′′(t)T (t)

=

constant = a





Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)


X′′(x)X(x)

= µ0ε0T ′′(t)T (t)

= constant =

a





Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)


X′′(x)X(x)

= µ0ε0T ′′(t)T (t)

= constant = a





Basic Examples

Newton’s Second Law.

md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation

for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics,

here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t)

is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time.

In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t .

Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force

bring us immediately to solving anODE.



Basic Examples


md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.



Basic Examples

Consider the Gravity near earth’s surface

md2rdt2 = −mgk

That vector equation is equivalent to the next three ODE’s

md2xdt2 = 0

md2ydt2 = 0

md2zdt2 = −mg



Basic Examples

Example 1.1

Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 for x > 0 .

SolutionWe will need the first and second derivatives :

y ′(x) = −32x−5/2 , y ′′(x) = 15

4 x−7/2

Plug these as well as the function into the differential equation:



Basic Examples

Example 1.1Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 for x > 0 .

SolutionWe will need the first and second derivatives :

y ′(x) = −32x−5/2 , y ′′(x) = 15

4 x−7/2

Plug these as well as the function into the differential equation:



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So,

y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2

does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy

the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation

andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence

is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution.

Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include

the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition

thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ?

I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use

this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere

in the workshowing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the work

showing that the function would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function

would satisfy the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy

the differential equation.



Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.



Basic Examples

To see why recall that

y(x) = x−3/2 = 1x3/2

In this form it is clear that we will need to avoid x = 0 at theleast as this would give division by zero.

So, we saw in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2

In this form

it is clear that we will need to avoid x = 0 at theleast as this would give division by zero.




Basic Examples


y(x) = x−3/2 = 1x3/2





Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw

in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw in the last example that even though a function maysymbolically satisfy

a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw in the last example that even though a function maysymbolically satisfy a differential equation,

because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution

we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues

of the independent variable and hence, must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2


So, we saw in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence,

must make arestriction on the independent variable.



Basic Examples


y(x) = x−3/2 = 1x3/2





Basic Examples

In the last example,

note that there are in fact many morepossible solutions to the differential equation. For instance all ofthe following are also solutions

y(x) = x−1/2

y(x) = 5x−3/2

y(x) = 3x−1/2

y(x) = 2x−1/2 + 3x−3/2



Basic Examples

In the last example, note that there are in fact many morepossible solutions

to the differential equation. For instance all ofthe following are also solutions

y(x) = x−1/2

y(x) = 5x−3/2

y(x) = 3x−1/2

y(x) = 2x−1/2 + 3x−3/2



Basic Examples

In the last example, note that there are in fact many morepossible solutions to the differential equation.

For instance all ofthe following are also solutions

y(x) = x−1/2

y(x) = 5x−3/2

y(x) = 3x−1/2

y(x) = 2x−1/2 + 3x−3/2



Basic Examples

In the last example, note that there are in fact many morepossible solutions to the differential equation. For instance all ofthe following are also solutions

y(x) = x−1/2

y(x) = 5x−3/2

y(x) = 3x−1/2

y(x) = 2x−1/2 + 3x−3/2



Basic Examples

I’ll leave the details to you to check that these are in factsolutions. Given these examples can you come up

with anyother solutions to the differential equation?

y(x) = c1x−1/2 + c2x−3/2

There are in fact an infinite number of solutions to thisdifferential equation.

So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples

I’ll leave the details to you to check that these are in factsolutions. Given these examples can you come up with anyother solutions to the differential equation?

y(x) = c1x−1/2 + c2x−3/2





Basic Examples


y(x) = c1x−1/2 + c2x−3/2

There are in fact

an infinite number of solutions to thisdifferential equation.




Basic Examples


y(x) = c1x−1/2 + c2x−3/2

There are in fact an infinite number of solutions

to thisdifferential equation.




Basic Examples


y(x) = c1x−1/2 + c2x−3/2





Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So,

given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given

that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are

an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number

of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions

to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation

in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example,

we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask

a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion.

Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is

the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution

that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or

does it matterwhich solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matter

which solution we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution

we use? This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use?

This question leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question

leads us to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us

to the nextmaterial in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2


So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial

in this section.



Basic Examples


y(x) = c1x−1/2 + c2x−3/2





Basic Examples

Example 1.2

Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 with initial conditions y(4) = 1

8 andy ′(4) = − 3

64 .

SolutionAs we saw in previous example the function y(x) = x−3/2 is asolution and we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64

and so this solution also meets the initial conditions.



Basic Examples

Example 1.2Show that y(x) = x−3/2

is a solution to4x2y ′′ + 12xy ′ + 3y = 0 with initial conditions y(4) = 1

8 andy ′(4) = − 3

64 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples

Example 1.2Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0

with initial conditions y(4) = 18 and

y ′(4) = − 364 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples

Example 1.2Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 with initial conditions

y(4) = 18 and

y ′(4) = − 364 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples

Example 1.2Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 with initial conditions y(4) = 1

8 and

y ′(4) = − 364 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .

Solution

As we saw in previous example the function y(x) = x−3/2 is asolution and we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .

SolutionAs we saw in previous example

the function y(x) = x−3/2 is asolution and we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .

SolutionAs we saw in previous example the function y(x) = x−3/2

is asolution and we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .

SolutionAs we saw in previous example the function y(x) = x−3/2 is asolution and

we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples


8 andy ′(4) = − 3

64 .


y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64




Basic Examples

In fact, y(x) = x−3/2

is the only solution to this differentialequation that satisfies these two initial conditions.

Initial Value ProblemAn Initial Value Problem (or IVP) is a differential equation alongwith an appropriate number of initial conditions.

Example 1.3The following is an example of IVP

4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

In fact, y(x) = x−3/2 is the only solution

to this differentialequation that satisfies these two initial conditions.



4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

In fact, y(x) = x−3/2 is the only solution to this differentialequation

that satisfies these two initial conditions.



4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

In fact, y(x) = x−3/2 is the only solution to this differentialequation that satisfies

these two initial conditions.



4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

In fact, y(x) = x−3/2 is the only solution to this differentialequation that satisfies these two

initial conditions.



4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

In fact, y(x) = x−3/2 is the only solution to this differentialequation that satisfies these two initial conditions.



4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value Problem

An Initial Value Problem (or IVP) is a differential equation alongwith an appropriate number of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value ProblemAn Initial Value Problem

(or IVP) is a differential equation alongwith an appropriate number of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value ProblemAn Initial Value Problem (or IVP)

is a differential equation alongwith an appropriate number of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value ProblemAn Initial Value Problem (or IVP) is a differential equation

alongwith an appropriate number of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value ProblemAn Initial Value Problem (or IVP) is a differential equation alongwith

an appropriate number of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples


Initial Value ProblemAn Initial Value Problem (or IVP) is a differential equation alongwith an appropriate number

of initial conditions.


4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples




4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples



Example 1.3

The following is an example of IVP

4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples



Example 1.3The following

is an example of IVP

4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples



Example 1.3The following is an example of

IVP

4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples




4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .



Basic Examples

Example 1.4

This is another example of an IVP

2ty ′ + 4y = 3, y(1) = −4 .

As you can notice, the number of initial conditions required willdepend on the order of the differential equation.

Interval of ValidityThe interval of validity for an IVP with initial condition(s)

y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples

Example 1.4This is

another example of an IVP

2ty ′ + 4y = 3, y(1) = −4 .



y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples

Example 1.4This is another example of

an IVP

2ty ′ + 4y = 3, y(1) = −4 .



y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples

Example 1.4This is another example of an IVP

2ty ′ + 4y = 3, y(1) = −4 .



y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .

As you can notice,

the number of initial conditions required willdepend on the order of the differential equation.


y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .

As you can notice, the number of initial conditions required

willdepend on the order of the differential equation.


y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .



y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .


Interval of Validity

The interval of validity for an IVP with initial condition(s)

y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .


Interval of ValidityThe interval of validity

for an IVP with initial condition(s)

y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .


Interval of ValidityThe interval of validity for an IVP

with initial condition(s)

y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples


2ty ′ + 4y = 3, y(1) = −4 .



y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.



Basic Examples

is

the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.

General SolutionThe general solution to a differential equation is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions into account.



Basic Examples

is the largest

possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval

on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which

the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution

is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid

andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 .

These are easy to define, but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define,

but can be very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be

very difficultto find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficult

to find in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find

in the practice.




Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.




Basic Examples


General Solution

The general solution to a differential equation is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution

to a differential equation is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation

is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation is the mostgeneral form

that the solution can take and doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation is the mostgeneral form that the solution

can take and doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation is the mostgeneral form that the solution can take and

doesn’t take anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation is the mostgeneral form that the solution can take and doesn’t take

anyinitial conditions into account.



Basic Examples


General SolutionThe general solution to a differential equation is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions

into account.



Basic Examples





Basic Examples

Example 1.5

y(t) = 34 + c

t2 is the general solution to the equation

2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c

t2

is the general solution to the equation

2ty ′ + 4y = 3




Basic Examples

Example 1.5

y(t) = 34 + c

t2 is the general solution

to the equation

2ty ′ + 4y = 3




Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3




Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!!

In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact,

all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions

to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equation

will be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form.

This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one

of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equations

that we will learn how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn

how to solve and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve

and we will be able to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able

to verify thisshortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify this

shortly.



Basic Examples

Example 1.5

y(t) = 34 + c


2ty ′ + 4y = 3




Basic Examples

Particular Solution

The particular solution to a differential equation is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).

Example 1.6What is the particular solution to the following IVP?

2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution

to a differential equation is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation

is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution

that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only

satisfies the differential equation, but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only satisfies the differential equation,

but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only satisfies the differential equation, but also

satisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given

initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).


2ty ′ + 4y = 3 y(1) = −4



Basic Examples


Example 1.6

What is the particular solution to the following IVP?

2ty ′ + 4y = 3 y(1) = −4



Basic Examples


Example 1.6What is

the particular solution to the following IVP?

2ty ′ + 4y = 3 y(1) = −4



Basic Examples


Example 1.6What is the particular solution

to the following IVP?

2ty ′ + 4y = 3 y(1) = −4



Basic Examples



2ty ′ + 4y = 3 y(1) = −4



Basic Examples

Solution

This is actually easier to do than it might, the general solution isof the form:

y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples

SolutionThis is actually

easier to do than it might, the general solution isof the form:

y(t) =34+

ct2


−4 =34+

c12



Basic Examples

SolutionThis is actually easier to do

than it might, the general solution isof the form:

y(t) =34+

ct2


−4 =34+

c12



Basic Examples

SolutionThis is actually easier to do than it might,

the general solution isof the form:

y(t) =34+

ct2


−4 =34+

c12



Basic Examples

SolutionThis is actually easier to do than it might, the general solution

isof the form:

y(t) =34+

ct2


−4 =34+

c12



Basic Examples

SolutionThis is actually easier to do than it might, the general solution isof the form:

y(t) =34+

ct2


−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need

is to determine the value of c that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine

the value of c that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c

that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give us

the solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution

that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after.

To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after. To find this,

all we need to do isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after. To find this, all we need to do

isuse our initial condition as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition

as follows:

−4 =34+

c12



Basic Examples


y(t) =34+

ct2


−4 =34+

c12



Basic Examples

c = −4− 34= −19

4

So, the particular solution to the IVP is:

y(t) =34− 19

4t2



Basic Examples

Implicit/Explicit Solution

An explicit solution is any solution that is given in the formy = y(t).

An implicit solution is any solution that is not in explicit form,that is, y can not be expressed explicitly as a function of t .Note that it is possible to have either general implicit/explicitsolutions and particular implicit/explicit solutions.

Example 1.7Show that y2 = t2 − 3 is an implicit solution of the differentialequation yy ′ = t .



Basic Examples

Implicit/Explicit SolutionAn explicit solution

is any solution that is given in the formy = y(t).





Basic Examples

Implicit/Explicit SolutionAn explicit solution is any solution

that is given in the formy = y(t).





Basic Examples

Implicit/Explicit SolutionAn explicit solution is any solution that is given in the formy = y(t).





Basic Examples


An implicit solution

is any solution that is not in explicit form,that is, y can not be expressed explicitly as a function of t .Note that it is possible to have either general implicit/explicitsolutions and particular implicit/explicit solutions.




Basic Examples


An implicit solution is any solution that is not in explicit form,that is,

y can not be expressed explicitly as a function of t .Note that it is possible to have either general implicit/explicitsolutions and particular implicit/explicit solutions.




Basic Examples


An implicit solution is any solution that is not in explicit form,that is, y can not be expressed explicitly as a function of t .Note that it is possible

to have either general implicit/explicitsolutions and particular implicit/explicit solutions.




Basic Examples


An implicit solution is any solution that is not in explicit form,that is, y can not be expressed explicitly as a function of t .Note that it is possible to have either general implicit/explicitsolutions and

particular implicit/explicit solutions.




Basic Examples






Basic Examples



Example 1.7

Show that y2 = t2 − 3 is an implicit solution of the differentialequation yy ′ = t .



Basic Examples



Example 1.7Show that y2 = t2 − 3 is an implicit solution

of the differentialequation yy ′ = t .



Basic Examples






Basic Examples

Solution

Using implicit derivation, the solution follows:

2yy ′ = 2t + 0

yy ′ = t

Example 1.8Find a particular explicit solution to the IVP

yy ′ = t y(2) = −1



Basic Examples

SolutionUsing implicit derivation,

the solution follows:

2yy ′ = 2t + 0

yy ′ = t


yy ′ = t y(2) = −1



Basic Examples

SolutionUsing implicit derivation, the solution follows:

2yy ′ = 2t + 0

yy ′ = t


yy ′ = t y(2) = −1



Basic Examples


2yy ′ = 2t + 0

yy ′ = t

Example 1.8

Find a particular explicit solution to the IVP

yy ′ = t y(2) = −1



Basic Examples


2yy ′ = 2t + 0

yy ′ = t


yy ′ = t y(2) = −1



Basic Examples

Solution

We already know from the previous example that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one will be the correct !!!We can determine the correct function by reapplying the initialcondition.



Basic Examples

SolutionWe already know

from the previous example that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples

SolutionWe already know from the previous example

that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples

SolutionWe already know from the previous example that an implicitsolution to this IVP is

y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples

SolutionWe already know from the previous example that an implicitsolution to this IVP is y2 = t2 − 3.

To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples

SolutionWe already know from the previous example that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution

allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples

SolutionWe already know from the previous example that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3




Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here.

There are two functions hereand we only want one and in factonly one will be the correct !!!We can determine the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and

in factonly one will be the correct !!!We can determine the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in fact

only one will be the correct !!!We can determine the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one

will be the correct !!!We can determine the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one will be the correct !!!

We can determine the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one will be the correct !!!We can determine

the correct function by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one will be the correct !!!We can determine the correct function

by reapplying the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in factonly one will be the correct !!!We can determine the correct function by reapplying

the initialcondition.



Basic Examples


y(t) = ±√

t2 − 3




Basic Examples

Only one of them will satisfy the initial condition.

In this case we can see that the negative solution, will be thecorrect one. The explicit solution is then

y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however that it will notalways be possible to find an explicit solution.



Basic Examples


In this case

we can see that the negative solution, will be thecorrect one. The explicit solution is then

y(t) = −√

t2 − 3




Basic Examples


In this case we can see that

the negative solution, will be thecorrect one. The explicit solution is then

y(t) = −√

t2 − 3




Basic Examples


In this case we can see that the negative solution,

will be thecorrect one. The explicit solution is then

y(t) = −√

t2 − 3




Basic Examples


In this case we can see that the negative solution, will be thecorrect one.

The explicit solution is then

y(t) = −√

t2 − 3




Basic Examples



y(t) = −√

t2 − 3




Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find

an explicit solution to thedifferential equation. It should be noted however that it will notalways be possible to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution

to thedifferential equation. It should be noted however that it will notalways be possible to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation.

It should be noted however that it will notalways be possible to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however

that it will notalways be possible to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however that it will notalways

be possible to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however that it will notalways be possible

to find an explicit solution.



Basic Examples



y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however that it will notalways be possible to find

an explicit solution.



Basic Examples



y(t) = −√

t2 − 3




Basic Examples

Example 1.9

A spherical raindrop evaporates at a rate proportional to itssurface area. Write a differential equation for the volume of theraindrop as a function of time.

Solution

If the volume at time t is denoted by V (t) and the surface attime t is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt

proportional to S(t)

dV (t)dt

= −αS(t)



Basic Examples

Example 1.9

A spherical raindrop evaporates at a rate proportional to itssurface area.

Write a differential equation for the volume of theraindrop as a function of time.

Solution


dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9

A spherical raindrop evaporates at a rate proportional to itssurface area. Write a differential equation

for the volume of theraindrop as a function of time.

Solution


dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution


dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t

is denoted by V (t) and the surface attime t is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t is denoted by V (t) and

the surface attime t is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t is denoted by V (t) and the surface attime t

is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t is denoted by V (t) and the surface attime t is denoted by S(t),

then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t is denoted by V (t) and the surface attime t is denoted by S(t), then the fact that raindrop evaporates

at a rate proportional to its surface area can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution

If the volume at time t is denoted by V (t) and the surface attime t is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area

can be write as

dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

Example 1.9


Solution


dV (t)dt


dV (t)dt

= −αS(t)



Basic Examples

were α is a constant of proportionality.

But, we have thefollowing relationship

V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3

where c is a constant.



Basic Examples

were α is a constant of proportionality. But, we have thefollowing relationship

V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒

r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) =

4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 =

− 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 =

− 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3

where c

is a constant.



Basic Examples


V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = − 4απ[3V4π

]2/3 = − 4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3




Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.

(a) Assuming that the drug is always uniformly distributedthroughout the bloodstream,write a differential equation for theamount of the drug that is present in the bloodstream at anytime.

(b) How much of the drug is present in the bloodstream after along time?



Basic Examples

Example 1.10

A certain drug

is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient.

Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug

enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h.

The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues

or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstream

at a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present,

with a rateconstant of 0.4(h)−1.





Basic Examples

Example 1.10






Basic Examples

Example 1.10


(a) Assuming that the drug

is always uniformly distributedthroughout the bloodstream,write a differential equation for theamount of the drug that is present in the bloodstream at anytime.




Basic Examples

Example 1.10


(a) Assuming that the drug is always uniformly distributed

throughout the bloodstream,write a differential equation for theamount of the drug that is present in the bloodstream at anytime.




Basic Examples

Example 1.10


(a) Assuming that the drug is always uniformly distributedthroughout the bloodstream,

write a differential equation for theamount of the drug that is present in the bloodstream at anytime.




Basic Examples

Example 1.10


(a) Assuming that the drug is always uniformly distributedthroughout the bloodstream,write a differential equation

for theamount of the drug that is present in the bloodstream at anytime.




Basic Examples

Example 1.10


(a) Assuming that the drug is always uniformly distributedthroughout the bloodstream,write a differential equation for theamount of the drug that is present

in the bloodstream at anytime.




Basic Examples

Example 1.10






Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation

Rate of change of Q(t) =

Rate at which Q(t) enters the bloodstream -

Rate at which Q(t) exits the bloodstream

dQdt

= drug entering − drug exitingdQdt

= (concentration)(rate of entering)− (concentration)(rate of exiting)



Basic Examples

Solution

(a) If the drug

is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed

throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream,

and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of

the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg

at time t isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t

isdenoted by Q(t), then the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then

the equation that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation

that we will use is abalance equation




dQdt





Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use

is abalance equation




dQdt





Basic Examples

Solution





dQdt





Basic Examples

Solution





dQdt

=

drug entering − drug exitingdQdt




Basic Examples

Solution





dQdt

= drug entering −

drug exitingdQdt




Basic Examples

Solution





dQdt

= drug entering − drug exiting

dQdt




Basic Examples

Solution





dQdt


=

(concentration)(rate of entering)− (concentration)(rate of exiting)



Basic Examples

Solution





dQdt


= (concentration)(rate of entering)− (concentration)(rate of exiting)Dr. Marco A Roque Sol Computational Neuroscience. Session 1-2


Basic Examples

dQdt

=

(5)(100)−Q(t)(0.4)

and the final equation is

dQdt

= 500−Q(t)(0.4)

(b)

Q′(t) = 500−Q(t)(0.4)

Q′(t)500−Q(t)(0.4)

= 1



Basic Examples

dQdt

= (5)(100)−Q(t)(0.4)

and the final equation is

dQdt

= 500−Q(t)(0.4)

(b)

Q′(t) = 500−Q(t)(0.4)

Q′(t)500−Q(t)(0.4)

= 1



Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t

0.4Thus, in the long run the amount of drug present is ... 500/0.4 =20 mg !!!



Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ =

ce−0.4t

Q(t) =500− ce−t




Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t




Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =

500− ce−t




Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t

0.4

Thus, in the long run the amount of drug present is ... 500/0.4 =20 mg !!!



Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t

0.4Thus, in the long run the amount of drug present is ...

500/0.4 =20 mg !!!



Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t



€¦ · ordinary differential equations definitions classification basic examples definitions...

Documents