# Homework Help: Linear differential equaton

1. Nov 13, 2005

### asdf1

for the following question:
y+2y-35y=12e^(5x)+37sin5x

my problem:
yh=c1e^(-7x)+c2e^(5x)
suppose yp=c3xe^(5x)+c4sinx+c5cos5x
then yp=c3e^(5x)+5c3xe^(5x)+5c4cos5x-5c5sin5x
so yp=8c3e^(5x)+(-60c4-10c5)sin5x+(-60c5+10c4)
which means that 8c3=12 => c3=2/3
also from (-60c4-10c5)=37 and (10c4-60c5)=0 implies that
c4=-3/5 and c5=-1/10
so y=c1e^(-7x)+c2e^(5x)+(3/2)xe^(5x) +(3/2)sin5x-(1/10)cos5x
but the correct answer should be
y=c1e^(-7x)+c2e^(5x)+xe^(5x) +(3/2)sin5x-(1/10)cos5x

i've checked this many times, but i don't know where my calculations went wrong... :P

2. Nov 13, 2005

### HallsofIvy

Since your coefficents are correct for sine and cosine, just recheck the exponential.
Let y= c3xe5x

Then y'= c3e5x+ 5c3xe5x just as you say.

y"= 5c3e5x+ (5c3e5x+ 25c3xe5x)
= 103e5x+ 25c3xe5x

I don't see where you got the "8".

Now y"= 103e5x+ 25c3xe5x
+2y'= 2c3e5x+10c3xe5x
-35y= -35c3xe5x

So we must have 12c3= 12.

3. Nov 13, 2005

### asdf1

thank you very much!!! i'll redo my calculations...