Is the back of the book wrong? Solving a Differential Equation with Cosh(x)

Sparky_
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Homework Statement



Solve : y'' -y = cosh(x)


Homework Equations





The Attempt at a Solution



the attempt 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_
 

Attachments

Physics news on Phys.org
The circled term is a solution to the homogeneous equation y''-y=0. It can be absorbed into the c1*exp(x)+c2*exp(-x).
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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