# Thread: Separable Differential Equation

1. ## Separable Differential Equation

$y'=2y^2lnx/x$

with the intial condition that y(1)=1. I tried to separate them and on the left side I have that 1/2y=(lnx)^2 over 2 plus C ... Am I on the right track?

2. $\int\frac{dy}{y^2}=\int\frac{2ln(x)}{x}$

$\frac{-1}{y}=(ln(x))^2+C_{1}$

$y=\frac{-1}{(ln(x))^2+C_{1}}$

3. I just went through it in my head, and using u=ln(x) to solve ln(x)/x i got the same as you, but you need to have a negative in there from solving int(y^-2), now you just need to use y(1)=1 to find C.