Question

If the Wronskian W of f and g is 3e^4t, and if f(t)=e^2t, find g(t). ( I got e^2t*g'(t)-(e^2t)'g(t)=3e^4t, e^2t*g'(t)-2e^2t*g(t)=3e^4t, g'(t)-2g(t)=3e^2t, p=-2, e^integral of p=e^-2t, now what?)

Answer 1

you have a form of \[g' -2g =3e^{2t}\] for convenience, I capitalize g to easy following. \[G'-2G=3e^{2t}\]
in this first order, we have p =-2 therefore \[\mu t = e^{\int (-2)dt}=e^{-2t}\]
\[and~e^{-\int {p (t)dt}}=e^{2t}\]
aply formula to find out G (or g(t))
\[g(t) = e^{2t}\int {3e^{2t}e^{-2t}}dt +C e^{2t}\]
\[g(t) = e^{2t}\int {3dt}+Ce^{2t}\]
\[g(t)= 3te^{2t}+Ce^{2t}=e^2t(3t+C)\]