1
$\begingroup$

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

  • 1
    I edited to improve formatting - hope I got it right.2011-07-08
  • 0
    @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
  • 0
    @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
  • 0
    Thanks, amWhy. Let's hope OP comes back to tell us what was really meant.2011-07-08
  • 0
    Yes, this $${e^{-t^2}\over(1-2t^2)^2}$$ is correct2011-07-08
  • 0
    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

2 Answers 2