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I am trying to solve the differential equation $$\frac{dy}{dx}=1+xy$$

I used the substitution $y=v+x$ so we get $$\frac{dy}{dx}=\frac{dv}{dx}+1$$ So the equations turns out to $$\frac{dv}{dx}+1=1+x(v+x)$$ $\implies$ $$\frac{dv}{dx}+(-x)v=x^2$$ whose Integrating factor is $e^{\frac{-x^2}{2}}$

So the solution is $$v \times e^{\frac{-x^2}{2}}=\int x^2e^{\frac{-x^2}{2}}dx+C$$

But i am unable to integrate the above integrand in terms of elementary functions.

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    $\begingroup$ The integral does not have an analytical solution, do you know what an error function is? $\endgroup$
    – Dhanvi Sreenivasan
    Commented Jun 10, 2022 at 11:49
  • $\begingroup$ Yes i know thank you i got it now $\endgroup$
    – Umesh shankar
    Commented Jun 10, 2022 at 11:51
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    $\begingroup$ Well the first equation itself was in the linear form where you could use the integrating factor method. $\endgroup$
    – Aman Kushwaha
    Commented Jun 10, 2022 at 13:12
  • $\begingroup$ @AmanKushwaha ya i know, but just tried in a different way $\endgroup$
    – Umesh shankar
    Commented Jun 10, 2022 at 13:32

1 Answer 1

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You can integrate the integrand if integrating over $]-\infty, \infty[$.

This is a classic Gaussian integral, which solves to $$ \int_{-\infty}^{\infty}x^2 e^{-x^2/2}dx = 2\int_{0}^{\infty}x^2 e^{-x^2/2}dx = \sqrt{2 \pi}. $$

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