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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. :)

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The Wronskian of f and g will be $e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$. When you divided out by $e^{2t}$, you forgot to divide the right hand side. This likely explains the the error (as the rest of the work looks fine). –  Aaron Nov 2 '11 at 19:08
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In your very first step, you forgot to divide $3e^{4t}$ by $e^{2t}$ as well. From the information given, you have: $$e^{2t}g'-2e^{2t}g=3e^{4t}$$ So dividing by $e^{2t}$ gives us the ODE $$g'-2g=3e^{2t}$$ The solution to this is as the book states.

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oh my gosh thank you. it sometimes takes a second set of eyes to catch something so simple as that. thank you soooooo much for that. –  AlexW.H.B. Nov 2 '11 at 19:08
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You know that $$W(f,g) = \left|\begin{array}{cc} f & g\\ f' & g' \end{array}\right| = \left|\begin{array}{cc} e^{2t} &g\\ 2e^{2t} & g' \end{array}\right| = e^{2t}g' - 2e^{2t}g = 3e^{4t}.$$

If you simplify by factoring out $e^{2t}$ and cancelling, that would give you $$g' - 2g = 3e^{2t},$$ instead of $g'-2g = 3e^{4t}$, which is what you had. So it looks like you made a simplification error.

Will this give you the correct answer?

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thank you so much for you answer! :) this has been so helpful. –  AlexW.H.B. Nov 2 '11 at 19:09
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