# Differential Equation Help

Discussion in 'Math' started by Sparky, Jun 19, 2007.

1. ### Sparky Thread Starter AAC Fanatic!

Aug 1, 2005
75
0
The problem is: Solve : y'' -y = cosh(x)

I've attempted a solution and is attached.

The answer in the back of the book agree with part of mine. However, the circled part - the part with 1/8 as a coefficient on e^x terms. It's circled on the bottom.

Either the back of the book is wrong (and I'm correct) or this term should cancel and I've missed something.
Help?

Thanks
-Sparky_

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2. ### AlexK Active Member

May 23, 2007
34
0
I didn't go through all your solution so i dont know if got it wrong or not.
this is how i solved it:
(1): y'' -y = cosh(x)=(exp(x)-exp(-x)/2

y=yh+yp;

yh=c*exp(-x)+c2exp(x);

then we guess a private solution: yp=k1*x*exp(x)+k2*x*exp(-x);
then we substitute this expression into (1) and find the constants k1,k2.
So the final solution is: y=yh+yp;

I solved it, and it matched the answer you said that was in the book.
Hope i helped.