I'm struggling with a seemingly simple problem in differential equations.
If the Wronskian W of $f$ and $g$ is $3e^{4t}$, and if $f(t) = e^{2t}$, find $g(t)$.
So from that I made a first order linear differential equation that is g'(t)-2g(t)=3e^{4t} then I proceeded to solve by using an integrating factor which is $e^{-2t}$ and what i got was $ g(t)=\frac{3}{2}(e^{4t})+Ce^{2t}$ but this is not the real answer according to my text book which says it is $g(t) = te^{2t} + ce^{2t}$ so my question is where am i going wrong? I don't understand where the $t$ came from in $te^{2t}$ any help would be greatly appreciated. :)