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I am looking for help with finding the integral of a given equation $$ Y2(t) = (1 - 2t^2)\int {e^{\int-2t dt}\over(1-2t^2)^2}. dt$$

anyone able to help? Thanks in advance!

UPDATE: I got the above from trying to solve the question below.

Solve, using reduction of order, the following $$y'' - 2ty' + 4y =0$$ , where $$f(t) = 1-2t^2$$ is a solution

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I edited to improve formatting - hope I got it right. – Gerry Myerson Jul 8 '11 at 2:06
@Gerry et.al. Note original code of OP: (e^(-t^2))/(1-2^(t^2)^2): it's 2 raised to the power $t^2$, but I'm not sure whether the exponent $t^2$ is squared, or what if it's (2 raised to $t^2$) all squared. – amWhy Jul 8 '11 at 2:30
@amWhy: That would make the denominator $1-2^{\left(t^4\right)}$, assuming exponentiation groups from right to left. I am sure Robert Israel's answer applies here as well. – Ross Millikan Jul 8 '11 at 2:49
Thanks, amWhy. Let's hope OP comes back to tell us what was really meant. – Gerry Myerson Jul 8 '11 at 3:05
Yes, this $${e^{-t^2}\over(1-2t^2)^2}$$ is correct – user10695 Jul 8 '11 at 4:27
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2 Answers

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There is no elementary antiderivative for this function. Neither Maple nor Mathematica can find a formula for it.

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What do you mean? It is actually one of the exercises from the class textbook. – user10695 Jul 8 '11 at 6:02
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Textbooks make mistakes. Robert doesn't. – Gerry Myerson Jul 8 '11 at 6:52
Hmm, interesting. The actual integral you want to do to solve this differential equation is $\int \frac{e^{t^2}}{(1-2 t^2)^2} \, dt$. The change from $-t^2$ to $t^2$ makes a big difference. This one can be done in closed form (though still not elementary): $\frac{\sqrt{\pi}}{4} \hbox{erfi}(t) - \frac{e^{t^2} t}{4 t^2 - 2}$. – Robert Israel Jul 8 '11 at 7:03
... but if this is a typical elementary differential equations course, I doubt that you'd be expected to come up with that. Just leave it in the form of an integral. – Robert Israel Jul 8 '11 at 7:10
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Oh, and I hate to contradict Gerry, but I do make mistakes. – Robert Israel Jul 8 '11 at 7:14
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The differential equation you have is a special case of the Hermite differential equation, with $\lambda =4$. The standard "regular" solution is the Hermite Polynomial $H_{\lambda/2}(x)=-2+4x^2$ (your solution is merely a scaled version), and the "irregular" solution is a bit complicated, involving the so-called "imaginary error function" $\mathrm{erfi}(x)$.

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