Rijad Hadzic
- 321
- 20
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}<br /> 0 & e^{-x} \\<br /> cosh(x) & -e^{-x} <br /> \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{<br /> \begin{vmatrix}<br /> e^x & 0 \\<br /> e^x & cosh(x)<br /> \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?