Calculus...?

Question

Solve the DE

(dy/dx) - y = e^(x) * y^(2).

Once again, I have already (attempted to) solve this, but am not sure whether correct or not! :)

Answer 1

y(x)=(-2e^(x)) / (e^(2x)-2C)

C being the constant, a constant mult by two is just a constant so you can make the 2C just a C.

It's correct, i checked by going pluging the funtion into the beginning ode equation and both sides equal eachother, that is a way you check yourself. So you should get an answer like that when fully simplified
if you can't figure it out and need the actual process instead of what the calc answer then message me and i'll get some paper and get back to you later.

Answer 2

(1 / y ²) dy/dx - 1 / y = e^x
Let z = - 1 / y = - y^(-1)
dz / dy = y ^(-2) = 1 / y ²
dz/dx = (dz/dy) (dy/dx) = (1 / y ²) (dy/dx)
dz/dx + z = e^x
Integrating factor = ∫ P dx = ∫1 dx = x
x z = ∫ x e^x dx
x z = x e^x - e^x + C
(- x / y) = x e^x - e^x + C
x / y = e^x - xe^x + K