# Could someone explain what I am missing in regard to this Wronskian?

The Wronskian of $f$ and $g$ is $t^2e^t$. If $f(t)=t$ what is $g$? Obviously, this reduces to the linear ODE $$g' + \dfrac1t g = t^2e^t$$ However, by tabular integration I arrive at a RHS of $e^tt^2 - 2te^t + 2e^t + c$.
But, the text asserts that this is equal to $te^t +ct$. How can the middle two terms collapse if the first one is multiplied by $t$ and second is not? That is, how is it that $t^2e^t - 2te^t + 2e^t = t^2e^t$?

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IMo the differential equation is $$tg'-g=t^2e^t$$ Where did you get that division by $\;t\;$ from? – DonAntonio Sep 30 '13 at 19:39
Correct, divide through now by t. This gives an integrating factor of e^(int(1/t)) = t – court Sep 30 '13 at 19:41
Even dividing by $\;t\;$ we get $$g'-\frac1tg=te^t$$Observe the sign in the left side and the exponent of $\;t\;$ in the right one. – DonAntonio Sep 30 '13 at 19:42
Hi, court, welcome on math.SE! I LaTeXed your equations and formulae. Please, have a look here and learn how to do it! :) – Andrea Orta Sep 30 '13 at 19:43
Will do. I know I need to learn the formatting. I'll make sure the next post is formatted correctly. – court Sep 30 '13 at 19:46

$$fg'-f'g=t^2e^t\;\;\wedge\;\;f(t)=t\;\implies\;tg'-g=t^2e^t\ldots$$
So that we indeed get $\;g(t)=te^t+ct\;$ , because
$$tg'-g=t(e^t+te^t+c)-te^t-ct=t^2e^t\ldots$$