Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$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?

share|improve this question
    
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

1 Answer 1

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 $C$.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.