Question

ydy + (xy²-8x) dx =0​

Answer 1

Explanation:

ydy+(xy^2-8x)dx=0

ydy + x(y^2 - 8)dx = 0

y/(8 - y^2) dy = x dx

Integrating both sides,

int(y(8-y^2)) dy = int( x) dx

-(1/2).ln(8 - y^2) = x^2/2 + c

ln(8 - y^2) = A -- x^2

8 - y^2 = e^(A - x^2)

y^2 = 8 - e^(A - x^2)