What is the general solution for a DE involving cosh and sinh?

[Rijad Hadzic]

Homework Statement

Solve the DE by variation of parameters:

$y'' - y = cosh(x)$

Homework Equations

The Attempt at a Solution

I got m = 1 and m = -1 so

$y = c_1e^x + c_2e^{-x} + y_p$

$y_p = u_1e^x + u_2e^{-x}$

The wonksian gave me -2

so

$u_1' = \frac{\begin{vmatrix} 0 & e^{-x} \\ cosh(x) & -e^{-x} \end{vmatrix} }{-2}$

I get $-\frac12 \int{e^{-x}cosh(x) dx} = u_1$

I do parts, u = e^-x and dv = coshx and eventually I get

$-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1$
then

$u_2' = \frac{ \begin{vmatrix} e^x & 0 \\ e^x & cosh(x) \end{vmatrix} }{-2}$

I get $-\frac12 \int{e^{x}cosh(x) dx} = u_2$

I do parts with u = e^x and dv = coshx and I get

$\frac 12 e^xsinh(x) + \frac 12 e^xcosh(x) = u_2$

$u_1*e^x = -\frac 14 sinh(x) + \frac 14 cosh(x)$
$u_2*e^{-x} = \frac 12 sinh(x) + \frac 12 cosh(x)$

adding I get $\frac 14 sinh(x) + \frac 34 cosh(x)$

so my general solution is

$y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)$

but my book got

$y_{gen} = c_1e^x +c_2e^{-x} + \frac 12 xsinh(x)$

how could it be possible to get a term called xsinh(x)??

This is the first time I've been given a problem with sinh and cosh. I've never been taught a single thing about them in school so I'm having trouble... I'm guessing there is an identity or something?

Would my answer of

$y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)$

satisfy the question?

 

[ehild]

Homework Statement

Solve the DE by variation of parameters:

$y'' - y = cosh(x)$

Homework Equations

The Attempt at a Solution

I got m = 1 and m = -1 so

$y = c_1e^x + c_2e^{-x} + y_p$

$y_p = u_1e^x + u_2e^{-x}$

The wonksian gave me -2

so

$u_1' = \frac{\begin{vmatrix} 0 & e^{-x} \\ cosh(x) & -e^{-x} \end{vmatrix} }{-2}$

I get $-\frac12 \int{e^{-x}cosh(x) dx} = u_1$

I do parts, u = e^-x and dv = coshx and eventually I get

$-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1$

You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(e^x+e^-x)

 

[Rijad Hadzic]

You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(e^x+e^-x)

Thank you I didn't even know this identity existed.

 

[ehild]

Thank you I didn't even know this identity existed.

http://www.sosmath.com/trig/hyper/hyper01/hyper01.html