Question

x³dy/dx + 3y² = xy²
Given x=1 when y=1

plz help me show step help!!

Answer 1

x^3 dy/dx + 3y^2 = xy^2
x^3 dy = y^2(x-3) dx
dy/y^2 = (x-3)/x^3 dx
-1/y = -1/x + 3/(2x^2) + c
1/y = (3-2x)/(2x^2) + c
y = 2x^2/(2x-3+cx^2)

y(1) = 1, so

2/(c-1) = 1
c = 3

y = 2x^2/(3x^2+2x-3)