y' = xe^(x+y)

[daiyo]

y' = xe^(x+y)
So I know you have to take the natural log of both sides.
ln y' = (x+y) ln x
Then I'm stuck.
Help please?
Thanks

 

[tkhunny]

So I know you have to take the natural log of both sides.

Why would you do that?

\(\displaystyle y' = x\cdot e^{x+y} = x\cdot e^{x}\cdot e^{y}\)

It's Separable!

 

[daiyo]

Oh ok so you would do:
S = integral sign
S(y'/e^y) = S (xe^x)

 

[tkhunny]

I think you have it. Let's see what you get.