Integrating Factor Method for Solving Differential Equations

[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

 

[Galileo]

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

 

[CarlB]

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

Carl

 

[Beretta]

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

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.

 

[Beretta]

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

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

 

[Tide]

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.

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

 

[saltydog]

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.

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?