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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 $t^2e^t - 2te^t + 2e^t = t^2e^t$?

Would someone please explain this to me? Thanks in advance.

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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
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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

1 Answer 1

up vote 1 down vote accepted

$$fg'-f'g=t^2e^t\;\;\wedge\;\;f(t)=t\;\implies\;tg'-g=t^2e^t\ldots$$

So that we indeed get $\;g(t)=te^t+ct\;$ , because

$$tg'-g=t(e^t+te^t+c)-te^t-ct=t^2e^t\ldots$$

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I don't understand that last line. –  court Sep 30 '13 at 19:49
    
Sorry but one does not usually solve an ODE by checking that some function fallen from the sky solves it... I am sure you can reformulate. –  Did Sep 30 '13 at 19:49
    
Okay, I am trying to figure out why -2t * e^t + 2*e^t = 0 –  court Sep 30 '13 at 19:54
    
I know it doesn't, @Did . Did you read the question? The OP says that is the answer but he can't understand why, so I corrected (I think...) his Wronskian and showed how what he says his text says is correct... –  DonAntonio Sep 30 '13 at 20:00
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Thank you, DonAntonio. I really appreciate it. –  court Oct 1 '13 at 15:12

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