1
$\begingroup$

$dy/dx = e^{-x^2} - 2xy $

$y(0) = 1$

expressing in the form $y = f(x)$

I was thinking of seperation of variable and the integrating factor method but I don't think it will work.

What should I do this?

  • 0
    Seperation of variable - no. But have you learned how to solve (and recognize) first-order linear equations?2012-05-19

1 Answers 1

4

WolframAlpha solves your problem quite perfectly:

$\frac{dy}{dx}+2xy=e^{-x^2}$

Multiplying by $e^{x^2}$:

$\frac{dy}{dx}e^{x^2}+(2e^{x^2}x)y=1$

The left side is $(e^{x^2}y)'$:

$(e^{x^2}y)'=1$

Integrate both sides with respect to $x$ and you have:

$e^{x^2}y(x)=x+C$

The only thing you need to do is using $y(0)=1$ to get rid of the constant. Put $x=0$ and $y=1$ to get $C$.

  • 0
    Yes, it was obviously a typo --- actually, two typos, one of which is still there. I understand "ping" to mean put your name with an at-sign in front, but that's unnecessary; as the author of the post to which I responded, you get notified whether I ping you or not. If you meant something else by "ping", I'm sorry; I plead old age. Anyway, I'll edit the other typo.2012-05-19