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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.2011-07-08
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    @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.2011-07-08
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    @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.2011-07-08
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    Thanks, amWhy. Let's hope OP comes back to tell us what was really meant.2011-07-08
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    Yes, this $${e^{-t^2}\over(1-2t^2)^2}$$ is correct2011-07-08
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    Then if, as Robert Israel says, there is no elementary antiderivative, you might want to check your work and see whether that's really the thing you wind up needing to integrate. If so, it just says that one of the solutions of the diff eqn can't be expressed in elementary terms.2011-07-08

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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.2011-07-08
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    Textbooks make mistakes. Robert doesn't.2011-07-08
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    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): $\frac{\sqrt{\pi}}{4} \hbox{erfi}(t) - \frac{e^{t^2} t}{4 t^2 - 2}$.2011-07-08
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    ... 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.2011-07-08
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    Oh, and I hate to contradict Gerry, but I do make mistakes.2011-07-08
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    Where did the negative sign go in your version? and yes, we are expected to solve the integral to obtain the second solution, Y2.2011-07-08
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    If $y = y_1$ is a solution of the second-order homogeneous linear differential equation $y'' + p y' + q y = 0$, then so is $y = y_1 u$ where $y_1 u'' + (2 y'_1 + p y_1) u' = 0$. In your case $p = -2 t$, $q = 4$, $y_1 = 1 - 2 t^2$, so the equation for $u$ is $(1 - 2 t^2) u'' + (-10 t + 4 t^3) u' = 0$, which we think of as a first-order linear equation in $u'$. Thus a solution is $$u' = \exp(\int - \frac{-10 t + 4 t^3}{1-2 t^2} \, dt) = \exp(t^2 - 2 \ln(-1 + 2 t^2)) = (1 - 2 t^2)^{-2} e^{t^2}$$ Just out of curiosity, what textbook is this?2011-07-08
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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" $\mathrm{erfi}(x)$.