# Initial-Value Problem

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

$y(0) = 1$

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? –  Gigili May 19 '12 at 9:08
Seperation of variable - no. But have you learned how to solve (and recognize) first-order linear equations? –  Gerry Myerson May 19 '12 at 9:21

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. –  JamesK May 19 '12 at 9:30
Okay, I'll explain it. –  Gigili May 19 '12 at 9:31
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? –  JamesK May 19 '12 at 9:32
@JamesK: I wrote a complete answer, see if you still have problems. –  Gigili May 19 '12 at 10:18
@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. –  Gerry Myerson May 19 '12 at 12:30