Question

If the Wronskian W of f and g is 3e^(4t) and if f(t)=e^(2t), find g(t).

Answer 1

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
\[g(t)=3te^{2t}+ke^{2t}\]

Answer 2

You are very detailed, and really really helped me understand what I was supposed to be doing. I thank you for that.