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

  • #1
Rijad Hadzic
321
20

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}
0 & e^{-x} \\
cosh(x) & -e^{-x}
\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{
\begin{vmatrix}
e^x & 0 \\
e^x & cosh(x)
\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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  • #2
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}
0 & e^{-x} \\
cosh(x) & -e^{-x}
\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)
 
  • #3
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.
 
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