Reminder siny = ex + C

This must be differentiated using implicit differentiation•When differentiating y’s write dy•When differentiating x’s write dx•Divide by dx

Implicit Differentiation

Implicit Differentiation

For example siny = ex + C

cosy dy = ex dx

dy

dxRearrange to make the subject

xdy e

dx cos y

xdycos y e

dx Divide by dx

Differentiating

siny = ex + C

cosy dy = ex dx

xdycos y e

dx

Reversing the Process

xdy e

dx cos y

xdycos y e

dx

cosy dy = ex dx

siny = ex + C

xcosy dy = e dx

mul

tiply

by

cosy

multiply by dxintegrate

Integrating

Finding the constant Csiny = ex + CTo find the constant C a boundary condition is needed.

If we are told that when x = 0 then y = /2 then we can find C.

siny = ex + C

Substitute x = 0 and y = /2

sin /2 = e0 + C

So C = 0 siny = ex

y = sin-1(ex)

1st Order Differential Eqns

Separating the Variables (C4)

1) Collect the x’s on one side and the y’s and the on the other side.

2) Multiply both side by dx and integrate.

3) Simplify the resulting equation.

y fns join the dy and x fns join the dx usually by Cross Multiplying

N.B y functions are integrated w.r.t y

x functions are integrated w.r.t x

Ex1 Solve the differential equation

Multiply both sides by y

Multiply both side by dx and integrate

Find C by substituting in 2 boundary conditions

2dy 3x

dx y

2dyy 3x

dx

2ydy 3x dx 2 3y 3x

C2 3

Differential equations where the variables cannot be separated

dy 3y1

dx x Ex1

If we try to rearrange so that the x`s and dx`s are on one side and the y`s and dy`s are on the other it is impossible.

Try it!!

dy 3y1

dx x

This is an example of a 1st order Linear Differential Equation.

It can be solved however by multiplying every thing by x3.

3 2 3dyx 3x y x

dx

What could we differentiate that would give this expression.

It must contain a mixture of x`s and y`s……..

3 2 3dyx 3x y x

dx

Differentiating x3y

x3dy + 3x2ydx using implicit differentiation divide by dx

3 2dyx 3x y

dx

So if we integrate3 2dy

x 3x ydx

we obtain x3y

3 2 3dyx 3x y x

dx

Integrate both sides

3 2 3dyx 3x y dx x dx

dx 4

3 xx y c

4

How do we find what to multiply by to make the LHS into an expression which can be integrated by inspection.

THE INTEGRATING FACTOR !!!!!!

dyP(x) y Q(x)

dx If

Then the integrating factor = P(x)dx

e

Differential equations where the variables cannot be separated

dyx 3y x

dx

Steps1. Rearrange in the form

2. Find the integrating factor I.F. =

3. Multiply every term by the I.F.

4. Integrate each side – the integral of the L.H.S. is I.F × y

dyP(x)y Q(x)

dx

P(x)dxe

dyx 3y x

dx

1. Rearrange in the form dy

P(x)y Q(x)dx

dy 3y1

dx x Divide by x

P(x) = and Q(x) = 1 3

x

dy 3y1

dx x

3. Multiply every term by the I.F.

3 2 3dyx 3x y x

dx

4. Integrate each side – the integral of the L.H.S. is I.F × y

3 2 3dyx 3x y dx x dx

dx 4

3 xx y c

4

dyP(x) y Q(x)

dx P(x) =

3

x

I.F = 3

dxP(x)dxxe e 3lnxe

3lnxe 3x

2. Find the integrating factor I.F. = P(x)dxe

How can you spot what the integral of the LHS is???

If we integrate3 2dy

x 3x ydx

we obtain x3y

This is the same as I.F. × y

So the integral of the LHS is I.F. × y

dy

dx+ cotx y  = cosecx x = 0 when y = 0

P(x) = cotx Q(x) = cosecxdy

P(x) y Q(x)dx

cot xdx lnsin xe e I.F = P(x)dx

e = sinx

dysin x

dx+ sinx cotx y  = sinx cosecx

1. Rearrange in the form dy

P(x)y Q(x)dx

2. Find the integrating factor I.F. = P(x)dx

e

3. Multiply every term by the I.F.

dysin x

dx+ sinx cotx y  = sinx cosecx

Simplify

dysin x

dx+ sinx y  = sinx

cos x

sinx

1

sinx

dysin x

dx+ cosx y  = 1

Integrating the LHS gives sinx y

sinx y = 1dx = x + c

i.e I.F × y

4. Integrate each side – the integral of the L.H.S. is I.F × y

sinx y = 1dx = x + c

Substitute x = 0 and y = 0 gives c = 0

sinx y = x So y = x

sin x