# Differential equation

1. Oct 22, 2006

### Punchlinegirl

Find the solution of the given inital value problem.
y"-2y'+y=te^t+4 y(0)=1 y'(0)=1

r^2-2r+1=0
(r-1)^2
so r= -1
my homogenous solution is y_h(t)=c_1e^-t+c_2te^-t

I have no idea where to begin for the particular solution. I know a need a guess, and I would think it would be Ate^t, but I really don't know.
Can someone please help me out?

2. Oct 22, 2006

### arildno

Is r=-1?
Why?

3. Oct 22, 2006

### Punchlinegirl

Oh sorry, I meant r= 1
then my homogenous solution would be c_1e^t +c_2te^t

4. Oct 22, 2006

### HallsofIvy

For a right hand side of the form
te^t+4
You should immediately think y= (At+ B)e^t+ C since when you have t^n, you might need lower powers of t.

Then, noticing that e^t and te^t are already solutions to the homogeneous equation, think of multiplying the e^t solution by the next power of t, t^2. That is, try y= (At^3+ Bt^2)e^t + C.

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