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Linear Algebra and Differential Equations
Sujin Khomrutai, Ph.D.
Department of Math & Computer ScienceChulalongkorn University
Lecture 1
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 1 / 31
Table of Contents
1 Introduction
2 Some Aspects
3 First Order ODEs: A Review
4 Second Order Linear Equations
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 2 / 31
What are DEs?
Definition
A differential equation (or DE) is an equation involving a function andsome of its derivatives.
EX.
1 y ′ = 2xy is a DE. (variable x , a function y(x).)
2 x2 + u2 = 4 is not a DE. No derivative!
3 t2 + (y ′′)2 = 4 is a DE. (variable t, a function y(t).)
4 ux + 5uy = 1 is a DE. (variables x , y , a function u(x , y).)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 3 / 31
What are DEs?
Definition
A system of DEs is a group of equations involving two or more functionsand some of their derivatives.
EX.
1
{u′ = 2u + v
v ′ = −u + 4v .⇒ a system of DEs. (functions u, v)
2
{x + 2y = 2
−x + y = 0.⇒ not a system of DEs. No derivatives!
3
u = 3u − v
v ′ = u + w
w ′ = −1
⇒ a system of DES. (functions u, v ,w)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 4 / 31
ODEs vs. PDEs
Definition
In a DE,
1 only one variable ⇒ ordinary differential equation (or ODE).
2 two or more variables ⇒ partial differential equations (or PDE).
System of ODEs = 1 variable, at least 2 functionsSystem of PDEs = at least 2 variables, at least 2 functions.
EX.
1 y ′ = 2xy is an ODE.
2 uxx + uyy = 1 is a PDE.
3
{u′ = 2u + v
v ′ = −u + 4vis a system of ODEs.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 5 / 31
Goal & Some Prep.
Goal. We study DEs and sytems of DEs.
Variable. t = time, or x = a space variable.
Functions. y (ODE), vectors ~x = (x1, . . . , xn) (Systems of ODES)
Derivatives.
y ′ =dy
dt, y ′′ =
d2y
dt2, . . . , y (n) =
dny
dtn, . . .
and~x ′ = (x ′1, x
′2, . . . , x
′n), ~x ′′ = (x ′′1 , x
′′2 , . . . , x
′′n ), . . .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 6 / 31
Some Aspects
Definition (Order)
For a DE or system of DEs, the highest order of derivative appeared iscalled the order.
EX.
1 y ′′ + 2y ′ = t3 ⇒ order = 2. (ODE)
2 y (4) − y ′′ + (y ′)5 = 0 ⇒ order = 4. (ODE)
3
{x ′1 = 2x1 − x2
x ′2 = −x1 + x2
⇒ order = 1. (system of ODEs)
4 ut + u(uxxt)5 = 0 ⇒ order = 3. (PDE)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 7 / 31
Some Aspects
Definition (Solutions)
A solution to a DE is a function that the equation becomes true when wesubstitute the function and its derivatives into the equation.
EX.
1 y ′ = 2t ⇒ y(t) = t2.
2 y ′ = 2y ⇒ y(t) = e2t .
3 y ′ + 3y = 6 ⇒ y(t) = e−3t + 2.
4 y ′′ − 3y ′ + 2y = 0 ⇒ y(t) = C1et + C2e
2t .
5 y (4) = 1 ⇒ y(t) = t4
24 .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 8 / 31
Some Aspects
Definition (Particular & General solutions)
A solution to a DE that contains no arbitrary constants is called aparticular solution.
A solution to a DE that contains some arbitrary constants and representsall possible particular solution is called a general solution.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 9 / 31
Some Aspects
EX. Consider the ODE y ′ = 2t.
y1(t) = t2 and y2(t) = t2 + 3 are solutions. They are particular solutions.
To find a general solution, we need integration
y ′ = 2t ⇒∫
y ′dt =
∫2tdt
which givesy(t)− y(0) = t2 ⇒ y(t) = t2 + C .
where C = y(0) is an arbitrary constant.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 10 / 31
Some Aspects
Definition (Linear vs Nonlinear)
An ODE is called a linear ODE if it can be expressed as
any(n) + · · ·+ a2y
′′ + a1y′ + a0y = f (t).
Otherwise, it is called nonlinear ODE.
EX.
1 y ′′ + 2ty ′ − 3y = et ⇒ linear.
2 ty (4) = 3y ′′ + sin t ⇒ linear.
3 y ′ + y2 = 0 ⇒ nonlinear.
4 y ′′′ + y ′′ + 2y ′ + 4y = x2 ⇒ linear. (Observe: variable is x .)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 11 / 31
Some Aspects
Definition (Initial Value Problems)
An initial value problem (or IVP) is a problem that has a DE togetherwith some conditions for the solutions at some t0 (or some x0).
Solving IVP ⇒ Particular solution.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 12 / 31
Some Aspects
EX. Solve the initial value problem
y ′ = 3t + 2, y(0) = 0.
Sol. We integrate
y ′ = 3t + 2 ⇒∫
y ′dt =
∫(3t + 2)dt
then
y(t)− y(0) =3t2
2+ 2t
Using y(0) = 0, we obtain
y(t) =3t2
2+ 2t.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 13 / 31
First Order ODE
Definition
A first order ODE is an equation of the form
F (t, y , y ′) = 0.
If it can be put into the form
y ′ = f (t)g(y)
it is called separable.
If it can be put into the form
y ′ + p(t)y = q(t)
it is a linear equation.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 14 / 31
Seprable Equations
How to solve separable eqns.?
To solve a separable equation, we write
dy
dt= f (t)g(y) ⇒ 1
g(y)dy = f (t)dt.
Then integrate ∫1
g(t)dy =
∫f (t)dt.
Solution can be implicitly defined.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 15 / 31
Separable Equations
EX. Solve the ODEdy
dt=
t
y.
Sol. We writeydy = tdt.
Integrate ∫ydy =
∫tdt ⇒ y(t)2
2− y(0)2
2=
t2
2.
Setting C = y(0)2, we get
y(t)2 = t2 + C .
This is a general solution.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 16 / 31
Separable Equations
EX. Sole the ODEdy
dx=
3x2
y − sin y.
Sol. We express
dy
dx=
3x2
y − sin y⇒ (y − sin y)dy = 3x2dx .
Then integrate∫(y − sin y)dy =
∫3x2dx ⇒ y2
2+ cos y = x3 + C .
The solution is implicit!
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 17 / 31
Separable Equations
EX. Sole the ODEy ′ = xy .
Sol. We write
y ′ = xy ⇒ 1
ydy = xdx .
Integrate ∫1
ydy =
∫xdx ⇒ ln y(t) =
x2
2+ C .
So
y(t) = ex2
2+C = De
x2
2 .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 18 / 31
First Order Linear ODE
Definition
For a linear equationy ′ + p(t)y = q(t),
the integrating factor is the function
I (t) = e∫p(t)dt .
Multiplying the ODE with I (t) and integrating, then
y(t) =1
I (t)
(∫I (t)q(t)dt + C
).
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 19 / 31
First Order Linear ODEs
EX. Solve the ODEy ′ + 3t2y = 6t2.
Sol. This is a linear equation: p(t) = 3t2 and q(t) = 6t2.
The integrating factor
I (t) = exp
∫3t2dt = et
3.
Plug into the formula
y(t) =1
et3
(∫et
36t2dt + C
)∴ y(t) = e−t
3(
2et3
+ C)
= 2 + Ce−t3.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 20 / 31
First Order Linear ODEs
EX. Solve the IVP (initial value problem)
x2y ′ + xy = 1 (x > 0), y(1) = 2.
Sol. Variable is x! Divide by x2 to bring coefficient of y ′ to 1:
y ′ +1
xy =
1
x2.
This is a linear equation: p(x) = 1/x and q(x) = 1/x2.
Integrating factor
I (x) = exp
∫1
xdx = e ln x = x .
Plug into the formula
y(x) =1
x
(∫x · 1
x2dx + C
)=
ln x + C
x.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 21 / 31
Second order ODEs
Definition
A second order ODE is an equation of the form
F (t, y , y ′, y ′′) = 0.
If the ODE can be put into the form
a(t)y ′′ + b(t)y ′ + c(t)y = f (t)
ory ′′ + p(t)y ′ + q(t)y = r(t)
it is called a second order linear ODE.
If a, b, c are constants, the ODE is called constant coefficients.
If f (t) = 0 or r(r) = 0, the ODE is called homogeneous.
Otherwise, it is nonhomogeneous.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 22 / 31
Second order ODEs
EX. Consider the ODE
3y ′′ + 2y ′ + 4y = t2 + 1.
This is a second order ODE.
It is linear, constant coefficients, and nonhomogeneous.
EX. Consider the ODE
5y ′′ + (tan t)y ′ + y2 = 0.
This is a second order ODE.
It is nonlinear because of the term y2.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 23 / 31
Second Order ODEs
EX. Consider the ODE
ty ′′ + (t + 1)y ′ + t2y = 0.
This is a second order ODE
a(t) = t, b(t) = t + 1, c(t) = t2, f (t) = 0.
It is linear, non-constant coefficients, and homogeneous.
We can divide by t to get another form
y ′′ +t + 1
ty ′ + ty = 0.
So p(t) = (t + 1)/t and q(t) = t.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 24 / 31
Initial Value Problems
Definition
If a second order ODE is supplied with two conditions of the form
y(t0) = k , y ′(t0) = l
it is called an initial value problem (or IVP).
Thus an IVP for a second order linear ODE is
(IVP)
{y ′′ + p(t)y ′ + q(t)y = f (t)
y(t0) = k , y ′(t0) = l .
Usually but not always, we choose t0 = 0.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 25 / 31
Existence and Uniqueness
Theorem
Consider an IVP of a second order linear ODE
(IVP)
{y ′′ + p(t)y ′ + q(t)y = f (t)
y(t0) = k , y ′(t0) = l .
If p(t), q(t), r(t) are continuous functions on an interval I , then the IVPhas a solution and it is unique.
The solution depends on the initial conditions k , l .
• The theorem can be applied to determine the longest interval I so thatthe problem has a unique solution.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 26 / 31
Existence and Uniqueness
EX. Find the longest interval I so that the IVP
(IVP)
{(t − 1)y ′′ + ty ′ + (t + 1)y = 0
y(2) = 2, y ′(2) = 3.
has a unique solution.
Sol. The conclusion does not involve the initial conditions!
Divide by (t − 1) so the coefficient of y ′′ becomes 1
y ′′ +t
t − 1y ′ +
t + 1
t − 1y = 0.
The functions p(t) = t/(t − 1), q(t) = (t + 1)/(t − 1), and r(t) = 0 arecontinuous on (−∞, 1) ∪ (1,∞).
The interval containing the initial time t0 = 2 is (1,∞) so the longestinterval is I = (1,∞).
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 27 / 31
Existence and Uniqueness
EX. Find the solution to the IVP
(IVP)
{y ′′ + p(t)y ′ + q(t)y = 0
y(t0) = 0, y ′(t0) = 0.
where p(t), q(t) are continuous functions.
Sol. By the theorem there is a unique solution.
Note that if we substitute y(t) = 0, it works
0′′ + p(t)0′ + q(t)0 = 0, 0(t0) = 0, 0′(t0) = 0.
So we get the solution to the IVP to be y(t) = 0.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 28 / 31
Principle of Superposition
Theorem
If y1, y2 are solutions of a homogeneous ODE
y ′′ + p(t)y ′ + q(t)y = 0,
then C1y1 + C2y2 is also a solution.
Note. It is not true for nonhomogeneous equation
y ′′ + p(t)y ′ + q(t)y = r(t).
In order that C1y1 + C2y2 is a general solution y1, y2 have to be extra!
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 29 / 31
Wronskian
EX. Consider the ODE
y ′′ + 3y ′ + 2y = 0.
It can be shown that
y1(t) = e−t , y2(t) = e−2t
are solutions. Using the superposition principle, we form a solution formula
y(t) = C1e−t + C2e
−2t .
Whether this is a general solution depends on whether it can satisfy theinitial conditions
y(t0) = k , y ′(t0) = l ,
for any k , l .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 30 / 31
Wronskian
Plug the formula y(t) into the conditions{C1y1(t0) + C2y2(t0) = k ,
C1y′1(t0) + C2y
′2(t0) = l .
This system can be solved for any k, l if and only if
det
(y1(t0) y2(t0)y ′1(t0) y ′2(t0)
)6= 0.
Definition
For any two functions f , g , the Wronskian of f , g is the function
W (f , g) = det
[f (t) g(t)f ′(t) g ′(t)
].
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 1 31 / 31