Question

The solution of $\frac{dy}{dx} = \frac{y^{3} + 2x^{2}y}{x^{3} + 2xy^{2}}$ is

• A. $(x^{2} - y^{2})^{3} = Bx^{2}y^{2}$
• B. $(x^{2} + y^{2})^{3} = Bx^{2}y^{2}$
• C. $(x^{2} - y^{2})^{3} = x^{2}y^{2}$
• D. None of these

Answer 1

Answer: (A)

$(x^{2} - y^{2})^{3} = Bx^{2}y^{2}$

Answer 2

Given equation is homogeneous. Let y = vx

∴ $\frac{dy}{dx} = v + x\frac{dv}{dx}$

⇒ $\frac{y^{3} + 2x^{2}y}{x^{3} + 2xy^{2}} = v + x\frac{dv}{dx}$ ⇒ $\frac{(y/x)^{3} + 2(y/x)}{1 + 2(y/x)^{2}} = v + x\frac{dv}{dx}$

⇒ $\frac{v^{3} + 2v}{1 + 2v^{2}} = v + x\frac{dv}{dx}$ ⇒ $x\frac{dv}{dx} = v\left\{ \frac{v^{2} + 2}{1 + 2v^{2}} - 1 \right\} = v\left\{ \frac{1 - v^{2}}{1 + 2v^{2}} \right\}$

⇒ $\frac{1 + 2v^{2}}{v(1 - v^{2})}dv = \frac{dx}{x}$ ⇒ $\frac{1 + 2v^{2}}{v(1 - v)(1 + v)}dv = \frac{dx}{x}$

⇒ $\left( \frac{A}{v} + \frac{B}{1 - v} + \frac{D}{1 + v} \right)dv = \frac{dx}{x}$

where $A(1 - v)(1 + v) + Bv(1 + v) + Dv(1 - v) = 1 + 2v^{2}$Putting

v = 0, A = 1

v = 1, $B = \frac{3}{2}$

v = – 1, $D = - \frac{3}{2}$

∴ $\left( \frac{1}{v} + \frac{3}{2}\frac{1}{1 - v} - \frac{3}{2}\frac{1}{1 + v} \right)dv = \frac{dx}{x}$

Integrating both side, we get

$\ln v + \frac{3}{2}\frac{\ln(1 - v)}{- 1} - \frac{3}{2}\ln(1 + v) = \ln x + \ln c$

⇒ $\ln v - \frac{3}{2}\ln(1 - v) - \frac{3}{2}\ln(1 + v) = \ln cx$ ⇒ $v/\{(1 - v)1 + v\}^{3/2} = cx$ ⇒ $\left( \frac{v}{cx} \right)^{2} = (1 - v^{2})^{3}$ ⇒ $\left( \frac{y}{cx^{2}} \right)^{2} = \left( 1 - \frac{y^{2}}{x^{2}} \right)^{3}$

⇒ $(x^{2} - y^{2})^{3} = \frac{x^{2}y^{2}}{c^{2}}$

∴ $(x^{2} - y^{2})^{3} = Bx^{2}y^{2}$, $\left( \because\frac{1}{c^{2}} = B \right)$