# 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?