Question

Solve the differential equation: x² y dx - (x³ + y³) dy = 0

Answer 1

Therefore the solution of given differential equation is $C=ye^{-\frac{x^3}{3y^3}}$

Given : The differential equation is x²y dx - (x³ + y³)dy = 0

To find : solve the given differential equation.

solution : x²y dx - (x³ + y³)dy = 0

⇒x²y/x³ dx - (x³/x³ + y³/x³)dy = 0

⇒(y/x) dx - {1 + (y/x)³} dy = 0

⇒(y/x) - {1 + (y/x)³}dy/dx = 0

let y/x = v ......(1)

now, y = vx

differentiating both sides w.r.t x we get,

dy/dx = v + x dv/dx

so, (y/x) - {1 + (y/x)³}dy/dx = 0

⇒(y/x) = {1 + (y/x)³} dy/dx

⇒v = (1 + v³)(v + x dv/dx)

⇒v = v + v⁴ + x dv/dx + v³x dv/dx

⇒0 = v⁴ + (1 + v³)x dv/dx

⇒- v⁴ dx = (1 + v³) x dv

⇒-∫dx/x = ∫(1 + v³)/v⁴ dv

⇒-lnx + lnC = v¯³/-3 + lnv

⇒ln(C/vx) = -v¯³/3

⇒C = (vx)e^{-v¯³/3}

putting, v = y/x

now C = (y/x × x)e^{-x³/3y³} = ye^{-x³/3y³}

Therefore the solution of given differential equation is $C=ye^{-\frac{x^3}{3y^3}}$

Answer 2

Answer:

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Step-by-step explanation: SOLVED BY USING Y=VX,