Diff Eq again?

Question

Can someone please help me solve this eqn? Thanks...

dy = x - 1 + xy - y

I've tried separing the variables all kinds of ways and keep getting dead-ended... please help!

Thank you... I didn't even consider factoring like that!!

Answer 1

dy/dx = x - 1 + xy - y = 1(x-1) +y(x-1) = (x-1)(y+1)
Separate the variables:
dy/(y+1) =(x-1) dx.
Integrating the two sides, we get:
ln(y+1) = (x-1)^2 /2 + c, where c = constant
It can be written as :
y+1 = A exp[(x-1)^2/2] , where A= exp(c) = constant. Or
y = A exp[(x-1)^2/2] -1 Answer
Verification : dy/dx = A exp[(x-1)^2/2] (x-1) = (y+1) (x-1)
= x-1 +xy -y OK!

Answer 2

I believe that the previous poster made a small mistake in the math; (has a "+x" instead of a "-x"). the correct solution is:

dy =(x-1)(y+1)dx,
ln(y+1)=1/2 x^2-x,
y=e^(1/2 x^2-x) - 1

Answer 3

dy =(x-1)(y+1)dx,
ln(y+1)=1/2 x^2+x,
y=e^(1/2 x^2+x) - 1