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

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Rijad Hadzic
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Homework Statement


Solve the DE by variation of parameters:

[itex]y'' - y = cosh(x)[/itex]

Homework Equations

The Attempt at a Solution


I got m = 1 and m = -1 so

[itex]y = c_1e^x + c_2e^{-x} + y_p[/itex]

[itex]y_p = u_1e^x + u_2e^{-x}[/itex]

The wonksian gave me -2

so

[itex]u_1' = \frac{\begin{vmatrix}<br /> 0 & e^{-x} \\<br /> cosh(x) & -e^{-x} <br /> \end{vmatrix} }{-2}[/itex]

I get [itex]-\frac12 \int{e^{-x}cosh(x) dx} = u_1[/itex]

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

[itex]-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1[/itex]
then

[itex]u_2' = \frac{<br /> \begin{vmatrix}<br /> e^x & 0 \\<br /> e^x & cosh(x)<br /> \end{vmatrix} }{-2}[/itex]

I get [itex]-\frac12 \int{e^{x}cosh(x) dx} = u_2[/itex]

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

[itex]\frac 12 e^xsinh(x) + \frac 12 e^xcosh(x) = u_2[/itex]

[itex]u_1*e^x = -\frac 14 sinh(x) + \frac 14 cosh(x)[/itex]
[itex]u_2*e^{-x} = \frac 12 sinh(x) + \frac 12 cosh(x)[/itex]

adding I get [itex]\frac 14 sinh(x) + \frac 34 cosh(x)[/itex]

so my general solution is

[itex]y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)[/itex]

but my book got

[itex]y_{gen} = c_1e^x +c_2e^{-x} + \frac 12 xsinh(x)[/itex]

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

[itex]y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)[/itex]

satisfy the question?
 
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Rijad Hadzic said:

Homework Statement


Solve the DE by variation of parameters:

[itex]y'' - y = cosh(x)[/itex]

Homework Equations

The Attempt at a Solution


I got m = 1 and m = -1 so

[itex]y = c_1e^x + c_2e^{-x} + y_p[/itex]

[itex]y_p = u_1e^x + u_2e^{-x}[/itex]

The wonksian gave me -2

so

[itex]u_1' = \frac{\begin{vmatrix}<br /> 0 & e^{-x} \\<br /> cosh(x) & -e^{-x}<br /> \end{vmatrix} }{-2}[/itex]

I get [itex]-\frac12 \int{e^{-x}cosh(x) dx} = u_1[/itex]

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

[itex]-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1[/itex]
You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(ex+e-x)
 
ehild said:
You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(ex+e-x)

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