Since the Wronskian is
[tex]W(f,g)=fg'-gf'=3e^{4t}[/tex]
and
[tex]f(t)=e^{2t}\text{ AND } f'(t)=2e^{2t}[/tex]
then
[tex]g'e^{2t}-2ge^{2t}=3e^{4t}[/tex]
Divide by [tex]e^{2t}[/tex]
[tex]g'-2g=3e^{2t}[/tex]
This is a nonhomogenous first order differential equation.
We write this in the form
[tex]g'+pg=h[/tex]
There is a really easy way to solve this. We introduce a dummy factor called the integrating factor.
Let
[tex]u'(t)=u(t)p
\\
\\ug'+gu'=(gu)'=uh
[/tex]
and
[tex]g=\frac{\int uh dt}{u}[/tex]
The integrating factor is
[tex]u(t)=e^{\int- 2dt+c}
\\
\\u(t)=e^{-2t+c}[/tex]
Then,
[tex]g(t)=\frac{\int u(t)(3e^{2t})dt+k}{u(t)}=\frac{\int e^{-2t+c}3e^{2t}dt+k}{e^{-2t+c}}=\frac{3\int e^{c}dt+k}{e^{-2t+c}}=\frac{3te^c+k}{e^{-2t+c}}[/tex]
So
[tex]g(t)=3te^{2t}+ke^{2t}[/tex]
