Math question...calculus?

Question

Please help me with this one. you have to solve the differential equation.

1. dy/dx=( y^2 + xy^2) / (x^2y - x^2)

Answer 1

dy/dx =y^2(1+x) /(x^2*( y-1))

dy/dx = (y^2/(y-1))*(1+x)/x^2

time dx both sides
dy = (y^2/(y-1))*(1+x)/x^2 dx
(y-1)dy/ y^2 = (1+x)dx/x^2
take integral both sides

int (y-1)dy/ y^2 = int (1+x)dx/x^2

figure out separate integral then put it back

int((y-1)dy/ y^2) = int (y/y^2 -1/y^2) dy
int(1/y -y^-2) dy
ln|y |+ y^-1
ln|y| + 1/y


int (1+x)dx/x^2
int (1/x^2 +x/x^2)dx
int (x^-2 +1/x)dx

-x^-1 +ln |x| +C
-1/x +ln|x|+C

combine together
ln|y| + 1/y = -1/x +ln|x|+C

Answer 2

Separation of variables

dy/dx = y^2 * (1+x) / [x^2 (y - 1)]
(y-1)/y^2 * dy = (1+x)/x^2 * dx
(1/y - 1/y^2) dy = (1/x^2 + 1/x) * dx

Now integrating
lny + 1/y = lnx - 1/x + constant

Or in a more elegant form:
ln|C*y/x| + (y+x)/(xy) = 0

Answer 3

dy/dx=( y^2 + xy^2) / (x^2y - x^2), so

(y x^2 - x^2) dy = (y^2 + x y^2) dx

(y - 1) x^2 dy = (1 + x) y^2 dx

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

1/y dy - 1/y^2 dy = 1/x^2 dx + 1/x dx

and these are all elementary integrals

Answer 4

dy/dx = y^2(1+x) / [x^2(y-1)]

((y-1)/y^2)dy = ((1+x)/(x^2))dx
[1/y - y^(-2) ]dy = [x^(-2) + 1/x] dx

Integrate both sides

ln|y| + 1/y = -1/x + ln|x| + c

Answer 5

y^2dx+xy^2dx=x^2ydy-x^2dy

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


∫dx/x^2+ ∫dx/x=∫dy/y-∫dy/y^2

The solution is
-1/x + ln x=lny + 1/y +C