I'm working through some Bernoulli differential equation exercises and wanted to check my work on one and get some guidance on one that is totally stumping me.
1.) $u' + u = e^xu^4$
Answer:
First, divide both sides by $u^4$, to get $u^{-4}u' + u^{-3} = e^x$
Now, substitute $v = u^{-3}$, and thus $v'= -3u^{-4}u'$
Plugging in and substituting, we get
$-\frac{1}{3}v' + v = e^x$, which means $v' - 3v = -3e^x$
Now we have a much nicer differential equation to work with, and,
$u(x) = e^{-\int-3}[\int e^{\int-3}(-3e^x)dx + c]$
$ = e^{3x}[\int e^{-3x}(-3e^x)dx + c]$
$ = e^{3x}[-3\int e^{-3x}e^xdx + c]$
$ = e^{3x}[-3\int e^{-2x}dx + c]$
$ = e^{3x}[-3(-\frac{1}{2}e^{-2x}) + c]$
$ = e^{3x}[\frac{3}{2}e^{-2x} + c]$ $= \frac{3}{2}e^x + ce^{3x}$
Correct?
2.) $xu' + u = xu'$
This is the one that is really stumping me from the start. I'm not sure where one would begin in order to work this into a simpler differential equation.