Question

x2 (y + 1) dx + y2 (x - 1) dy = 0.​

Answer 1

Step-by-step explanation:

y2(1−x)dydx=x2(y+1)

Dividing through, we have:

y2y+1dy=x21−xdx

The variables are separable, so we can integrate separately:

∫y2y+1dy=∫(u−1)2udu=∫u−2+1udu=u22−2u+lnu=

y2−2y+12+ln(y+1)+C

∫x21−xdx=∫(1−u)2udx=∫u−2+1udu=u22−2u+lnu=

x2−6x+12+ln(x+1)+C

So:

y2−2y+12+ln(y+1)=x2−6x+12+ln(x+1)+C

Solve the separable equation x^2 (y(x) + 1) + ( dy(x))/( dx) (x - 1) y(x)^2 = 0:

Solve for ( dy(x))/( dx):

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

Divide both sides by -(y(x) + 1)/y(x)^2:

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

Integrate both sides with respect to x:

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

Evaluate the integrals:

-log(y(x) + 1) - 1/2 (y(x) + 1)^2 + 2 (y(x) + 1) = x^2/2 + x + log(x - 1) + c_1 - 3/2, where c_1 is an arbitrary constant.

Simplify the arbitrary constants:

Answer: -log(y(x) + 1) - 1/2 (y(x) + 1)^2 + 2 (y(x) + 1) = x^2/2 + x + log(x - 1) + c_1