xy' + (1+x)y = e^(-x)sin2x
i am having trouble finding the integrating factor.
i got I.F = e^(1/2(1+x)^2) by using p(x)= 1+x and I.F = e^(integral 1+x)
but i was told I.F = xe^x
No, the DE is
$\displaystyle \displaystyle x\,y' + (1 + x)\,y = e^{-x}\sin{2x}$
$\displaystyle \displaystyle y' + \left(\frac{1 + x}{x}\right)y = x^{-1}e^{-x}\sin{2x}$
$\displaystyle \displaystyle y' + \left(\frac{1}{x} + 1\right)y = x^{-1}e^{-x}\sin{2x}$.
So the integrating factor is $\displaystyle \displaystyle e^{\int{\frac{1}{x} + 1\,dx}}$.
Well, the idea there is the whole idea behind the integrating factor method. You're trying to render the LHS of the DE a perfect derivative so you can just integrate. Use the product rule on $\displaystyle (xe^{x}y)'$ to get the other expression. As for going the other way, if you study the integrating factor method enough, you will see that that is the goal you're working towards. You'll learn to recognize perfect derivatives (integrable combinations is another name for them). That's part of solving DE's.