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