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I'm really stuck trying to verify that the given function is a solution of the differential equation.

I've attempted applying converting it to polar coordinates but I don't think I'm on the right track.

What are the first few steps that I need to take?

Thanks so much.

Integral of e^(-x^2)

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    $\begingroup$ Polar coordinates are totally irrelevant. You should differentiate the function $y$ suggested as a solution and compare the result to $2ty+1$. What is stopping you in the computation of $y'$, exactly? $\endgroup$ – Did Apr 9 '14 at 10:01
  • $\begingroup$ I was trying to help someone and had forgotten my basics and didn't know where to start. After I remembered the fundamental theorem of calculus I was able to do it. Thanks guys. :) $\endgroup$ – Will Apr 9 '14 at 10:40
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You have to apply the Fundamental Theorem of Calculus. For brevity, I set $I(t)=\int_0^te^{-s^2}ds$:

$$y'=2te^{t^2}I(t)+\underbrace{e^{t^2}e^{-t^2}}_{=1}+2te^{t^2}$$

hence

$$y'-2ty=2te^{t^2}I(t)+1+2te^{t^2}-2te^{t^2}I(t)-2te^{t^2}=1$$

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  • $\begingroup$ Thanks. After I remembered the fundamental theorem of calculus I was good to go. :) $\endgroup$ – Will Apr 9 '14 at 10:38

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