Maths
posted by Taldil on .
Please help me solve this differential equation: dy/dx(1+1/x)y=y^2

This is just a special case of the Bernoulli Equation
y' + p(x) y = q(x) y^n
where q(x) = 1
Make the substitution v = y^(1n)
This will give you
v' + (1+1/x)v = 1
Now you can multiply by an integrating factor
e^[∫(1+1/x) dx] = xe^x to get
xe^x v = e^x (x1)
and you're almost done 
Hi..
I'm not sure how to get the final answer