Solve the given initial-value problem. The DE is homogeneous. xy²dy/dx = y³ - x³; y(1) = 2.

Solution:

Given a homogenous differential equation, xy² dy/dx = y³ - x³

xy²dy/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,

⇒ dv/dx + v = (v³ - 1)/v²

⇒ dv/dx = [(v³ - 1)/v²] - v

⇒ dv/dx = (v³ - 1 - v³)/v²

v². dv = -1/x.dx

Integrate on both sides, we get

∫v². dv = ∫-1/x.dx

v³/3 = -logx + C

[1/3(y³/x³)] + log x = C --- (2)

Now, given that y(1) = 2

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

[(1/3) × (2³/1³)]+ log 1 = C

C = 8/3

From equation (2), we get

⇒ (y³/3x³) + logx = 8/3

Therefore, the required differential equation is y³ + 3x³logx = 8x³.

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Solve the given initial-value problem. The DE is homogeneous. xy²dy/dx = y³ - x³; y(1) = 2.

Summary:

The differential equation for the given initial-value problem xy²dy/dx = y³ - x³; y(1) = 1 is y³ + 3x³logx = 8x³.