# Thread: Help finding the integrating factor needed.

1. ## Help finding the integrating factor needed.

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

2. No, the DE is

$\displaystyle x\,y' + (1 + x)\,y = e^{-x}\sin{2x}$

$\displaystyle y' + \left(\frac{1 + x}{x}\right)y = x^{-1}e^{-x}\sin{2x}$

$\displaystyle y' + \left(\frac{1}{x} + 1\right)y = x^{-1}e^{-x}\sin{2x}$.

So the integrating factor is $\displaystyle e^{\int{\frac{1}{x} + 1\,dx}}$.

3. oh yes i see my mistake.. thanks.

so now for the D.E i have y'xe^x + (e^x + xe^x)y = sin2x

i forget how to simplify from here?

is it y'xe^x = sin2x - (e^x + xe^x)y
and then take the integral of both sides.. this looks messy

4. Originally Posted by linalg123
oh yes i see my mistake.. thanks.

so now for the D.E i have y'xe^x + (e^x + xe^x)y = sin2x

i forget how to simplify from here?
You can do

$(xe^{x}y)'=\sin(2x),$

and integrate.

5. can you show how i go from

y'xe^x + (e^x + xe^x)y to (xe^xy)'

6. 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 $(xe^{x}y)'$ 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.

7. ok so i got the R.S = -1/2cos(2x) but i am having trouble integrating (xe^xy)' any hints?

8. wait would it just be xe^x cause it's the integral of a derivitave?

9. It would be $\displaystyle x\,e^{x}y$, for the reason you have stated.

10. so then xe^xy = -(1/2)cos2x + c
y = -cos2x/(2(xe^x)) + c/(xe^x)

it then asks if there are any transient terms. that are the ones with exponentials right? so they would both be transient?

11. What do you mean by transient?

12. that's what the question asks.. I'm pretty sure it's the terms with e^x in them

13. Considering that all the terms die off at infinity, I'd say both terms on the RHS are transient. That's one definition that I've seen, anyway.