Solve the given initial-value problem. The DE is homogeneous. xy² dy/dx = y³ - x³; y(1) = 1.

Solution:

Given that, xy² dy/dx = y³ - x³

xy² dy/dx = y³ - x³

dy/dx = (y³ - x³)/ xy²

dy/dx = [(y/x)³-1]/ (y²/x²)............(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 = (v³-1)/v²

x. dv/dx = [(v³-1)/v²] - v

x. dv/dx = (v³-1-v³)/ v²

v². dv = -1/x .dx

v³/3 = -logx + C

[1/3 (y³/x³ )]+ log x = C …………………(2)

Now, given that y(1) = 1

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

[(1/3) × (1³/1³)]+ log 1 = C

C = 1/3

∴The required ‘Differential Equation' is,

(y³/3x³)+ logx= 1/3

y³+ 3x³logx =x³

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Solve the given initial-value problem. The DE is homogeneous. xy² dy/dx = y³ - x³; y(1) = 1.

Summary:

The ‘Differential Equation’ for the given initial-value problem xy² dy/dx = y³ - x³; y(1) = 1 is y³+ 3x³logx = x³.