Question

Solve using variation of parameters.
\[y''-4y=\frac{e^{2x}}{x}\]
\[y_1=e^{2x}, y_2=e^{-2x}, y_p=u_1y_1+u_2y_2\]
\[W = \begin{vmatrix} e^{2x} & e^{-2x}\\2e^{2x} & -2e^{-2x}
\end{vmatrix} = -4\]
\[W_1 = \begin{vmatrix} 0 & e^{-2x}\\ \frac{e^{2x}}{x} & -2e^{-2x}
\end{vmatrix} = -x^{-1}\]
\[W_2 = \begin{vmatrix} e^{2x} & 0\\2e^{2x} & \frac{e^{2x}}{x}
\end{vmatrix} = \frac{e^{4x}}{x}\]
\[u'_1 = \frac{W_1}{W}=\frac{1}{4x}, u_1=\frac{1}{4}\ln x\]
\[u'_2 = \frac{W_2}{W}=-\frac{e^{4x}}{4x},u_2=-\int \frac{e^{4x}}{4x} dx =?\]

Answer 1

Alrighty so this is differential equations. I haven't taken this in years, however, I was good at it when I took it. I think I might be able to help.
I would just need to look up variation of parameters in my book.

Answer 2

oh the Wronskian

Answer 3

Hmm I can't find any mistakes in your work....
That last integral is the `exponential integral`: noted by \(\Large\rm Ei(x)\)
It's not solvable using normal methods :o

Maybe uhhh power series or something... hmm thinking

Answer 4

so we need to integrate that

Answer 5

hm, the difficult part is the 1/4x part

Answer 6

because we know how to integrate e^4x

Answer 7

Have you tried integration by parts ? @heril

@zepdrix yep, looks like you're right hmm...

Answer 8

\[\Large\rm -\frac{1}{4}\int\limits \frac{e^{4x}}{x} dx =-\frac{1}{4}\int\limits \frac{1+4x+\frac{(4x)^2}{2!}+\frac{(4x)^3}{3!}+...}{x} dx\]You could write the power series for e^4x like that,
and then divide an x out of each term.
The first term ends up giving you ln(x) and then I bet after we integrate we can group the other terms somehow maybe :o hmmm

Answer 9

yep, not seeing an easy way to do that except specifying that it is the Expontial integral