Question

Solve dy/dx=(y³+3x²y) /(x³+3xy²)

Answer 1

In the last step substitute the value for v i.e. v = y/x

Answer 2

we have to solve dy/dx = (y³ + 3x²y)/(x³ + 3xy²)

dy/dx = (y³/x³ + 3x²y/x³)/(x³/x³ + 3xy²/x³)

= {(y/x)³ + 3(y/x)}/{1 + 3(y/x)²}

let y/x = P ⇒y = Px

differentiating both sides,

dy/dx = P + x dP/dx

after substitution we get,

P + x dP/dx = (P³ + 3P)/(1 + 3P²)

⇒x dP/dx = (P³ + 3P)/(1 + 3P²) - P

⇒x dP/dx = (P³ + 3P - P - 3P³)/(1 + 3P²)

⇒x dP/dx = (2P - 2P³)/(1 + 3P²)

⇒(1 + 3P²)dP/(2P - 2P³) = dx/x

⇒∫(1 + 3P²)dP/2P(1 - P)(1 + P)= ∫dx/x

using partial fraction method ,

(1 + 3P²)/P(1 - P)(1 + P) = A/P + B/(1 - P) + C/(1 + P)

1 + 3P² = A(1 - P²) + B(P + P²) + C(P - P²)

1 + 3P² = (B + C)P + (-A + B - C)P² + A

on comparing we get,

A = 1 , B + C = 0, -A + B - C = 3

-1 -C - C = 3 ⇒C = -2 and B = 2

(1 + 3P²)/2P(1 - P)(1 + P) = 1/2P + 1/(1 - P) - 1/(1 + P)

now, ∫dP/2P + ∫dP/(1 - P) - ∫dP/(1 + P) = ∫dx/x

⇒1/2 lnP + ln(1 - P) - ln(1 + P) = lnx + lnC

⇒ln{P(1 - P)²/(1 + P)²} = 2lnCx

⇒P(1 - P)²/(1 + P)² = C²x²

⇒(y/x)(1 - y/x)²/(1 + y/x)² = Kx²

⇒y(x - y)²/(x + y)² = Kx³

hence solution of equation is y(x - y)²/(x + y)² = Kx³