# Solve the given initial-value problem. The DE is homogeneous. xy^{2} dy/dx = y^{3} - x^{3}; y(1) = 1.

**Solution:**

Given that, xy^{2} dy/dx = y^{3} - x^{3}

xy^{2} dy/dx = y^{3} - x^{3}

dy/dx = (y^{3} - x^{3})/ xy^{2}

dy/dx = [(y/x)^{3}-1]/ (y^{2}/x^{2})............(1)

Put y/x = v ⇒ y = xv

Differentiate w.r.t x

dy/dx = x. dv/dx + v

∴ Equation (1) becomes,

x. dv/dx + v = (v^{3}-1)/v^{2}

x. dv/dx = [(v^{3}-1)/v^{2}] - v

x. dv/dx = (v^{3}-1-v^{3})/ v^{2}

v^{2}. dv = -1/x .dx

v^{3}/3 = -logx + C

[1/3 (y^{3}/x^{3} )]+ log x = C …………………(2)

Now, given that y(1) = 1

∴ Substituting y =1 and x =1 in equation(2)

[(1/3) × (1^{3}/1^{3})]+ log 1 = C

C = 1/3

∴The required ‘Differential Equation' is,

(y^{3}/3x^{3})+ logx= 1/3

y^{3}+ 3x^{3}logx =x^{3}

## Solve the given initial-value problem. The DE is homogeneous. xy^{2} dy/dx = y^{3} - x^{3}; y(1) = 1.

**Summary:**

The ‘Differential Equation’ for the given initial-value problem xy^{2} dy/dx = y^{3} - x^{3}; y(1) = 1 is y^{3}+ 3x^{3}logx = x^{3}.