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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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    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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    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" $\mathrm{erfi}(x)$.