Differential Equation

1. Dec 11, 2005

Beretta

Hi all,

Is the general solution of xy' + (1+x) y = e^(-x) sin 2x,
y= (-cos2x)/(2xe^(x)) + c/(xe^(x))

Thank you very much guys

2. Dec 11, 2005

Galileo

Why don't you just plug it in so you can check for yourself whether it is correct?

3. Dec 11, 2005

CarlB

Well the "c/(xe^(x))" part looks good.

Carl

4. Dec 11, 2005

Beretta

Because the other way arround is too long, and I'm not that good to garanty free-mistakes in the backward if I couldn't in the forward.

5. Dec 11, 2005

Beretta

I solved it again I got (-cos2x)/(2xe^(x)) that (1/2)(1/x)(cos2x/e^x)

6. Dec 11, 2005

Tide

If you won't do it "backward" then you'll never know if your "forward" is right. My advice: Just do it!

7. Dec 12, 2005

saltydog

Well, first write it as:

$$y^{'}+\frac{1+x}{x}y=\frac{Sin[2x]}{xe^x}$$

right?

Then solve for the integrating factor:

$$\sigma=Exp[\int (1+1/x)dx]$$

or:

$$\sigma=xe^x$$

multiplying both sides of the DE by this integrating factor results in:

$$d\left[xe^x y]=Sin[2x]dx$$

Can you finish it now?

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