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$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?

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    Since this is your third question, you've learned to use $\LaTeX$ I presume?2012-05-19
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    Seperation of variable - no. But have you learned how to solve (and recognize) first-order linear equations?2012-05-19

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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 $C$.

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    Not to sure, this is what I try using IFM.2012-05-19
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    Okay, I'll explain it.2012-05-19
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    dy/dx=e−x2−2xy dy/dx + 2xy = e^(-x^2) I let g(x) be 2xy so, P(x) = exp(integration of g(x)) Will give me, P(x) = exp(x^2) Right?2012-05-19
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    @JamesK: I wrote a complete answer, see if you still have problems.2012-05-19
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    I think you want $(e^{x^2}y)'=1$. Oh, and those $dx/dy$ terms are surely meant to be $dy/dx$.2012-05-19
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    @James, how is anyone supposed to know that when you write e-x2 you really mean $e^{-x^2}$? Can you learn a little TeX? All you need to do is write e^{-x^2}, but enclose it in dollar signs.2012-05-19
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    @GerryMyerson: You can simply edit or ping me instead of telling the OP what is meant by what. It's obvious that it was a typo anyway.2012-05-19
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    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