# Differential equations assignment T5

1. Dec 4, 2013

### mathi85

Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
Thank you in advance for your time.

Find the particular solution of the following differential equations:
a) 12(d2y/dx2)-3y=0
given that: x=0, y=3 and (dy/dx)=0.5

b) (d2y/dx2)+2(dy/dx)+2y=10ex
given that: x=0, y=0 and (dy/dx)=1

Solution:
a)
12(d2y/dx2)-3y=0 /:12
(d2y/dx2)-(1/4)y=0

m2=n2
m=+/-n

∴y=Aenx+Be-nx
or y=Acosh(nx)+Bsinh(nx)

m2=1/4
∴m=+/-√(1/4)=+/-(1/2)

y=Acosh[(1/2)x]+Bsinh[(1/2)x]

3=Acosh[0]+Bsinh[0]
3=A(e0+e0/2)+B(e0-e0/2)
A=3

dy/dx=Asinh(0)+Bcosh(0)
B=1/2

Solution
b)
d2y/dx2+2(dy/dx)+2y=10ex

m2+2m+2=0
∴m=-1+/-j

CF:
u=e-x{Acos(x)+Bsin(x)}

PI:
v=pex
v'=pex
v''=pex

pex+2pex+2pex=10ex
5pex=10ex /:5ex
p=10ex/5ex=2

v=2ex

GS:
y=e-x{Acos(x)+Bsin(x)}+2ex [1]

dy/dx=e-x{-Asin(x)+Bcos(x)}-e-x{Acos(x)+Bsin(x)}+2ex [2]

Sub into [1]
0=e0{Acos(0)+Bsin(0)}+2e0
A=-2

Sub into [2]
1=e0{2sin(0)+Bcos(0)}-e0{2cos(0)+Bsin(0)}+2e0
1=B-2+2
B=1

PS:
y=e-x{-2cos(x)+sin(x)}+2ex

Last edited: Dec 4, 2013
2. Dec 4, 2013

### HallsofIvy

Staff Emeritus
This is correct but I don't see why you would change to exponential form. If you are using hyperbolic functions you should know that a good reason for doing that is that cosh(0)= 1 and sinh(0)= 0. 3= Acosh(0)+ Bsinh(0) is immediately "3= A".

Yes, dy/dx(0)= B= 0.5
Now, how about actually writing the solution to the problem?

You've lost a sign here: the second term should be -e0(-2cos(0)+ Bsin(0))

So this should be 1= B+ 2+ 2

3. Dec 4, 2013

### mathi85

Thank you for the reply!

I'm actually not sure why I did it.

y=3cosh(x/2)+0.5sinh(x/2)

B=-3

∴y=e-x{-2cos(x)-3sin(x)}+2ex