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}<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?