Help finding the integrating factor needed.

**linalg123** · March 29th 2011, 05:13 AM

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

**Prove It** · March 29th 2011, 05:28 AM

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 $\displaystyle e^{\int{\frac{1}{x} + 1\,dx}}$ .

**linalg123** · March 29th 2011, 06:29 AM

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

**Ackbeet** · March 29th 2011, 06:34 AM

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.

**linalg123** · March 29th 2011, 06:50 AM

can you show how i go from

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

**Ackbeet** · March 29th 2011, 06:54 AM

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.

**linalg123** · March 29th 2011, 01:31 PM

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

**linalg123** · March 29th 2011, 03:00 PM

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

**Prove It** · March 29th 2011, 04:39 PM

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

**linalg123** · March 29th 2011, 04:44 PM

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?

**Prove It** · March 29th 2011, 04:46 PM

What do you mean by transient?

**linalg123** · March 29th 2011, 04:56 PM

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

**Ackbeet** · March 29th 2011, 05:57 PM

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.

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