Question

The solution of the differential equation $x{y}^{2}dy-(x^3+y^3)dx=0$ is

• A. $y^3=3x^3+c$
• B. $y^3=3x^3\log (cx)$
• C. $y^3=3x^3+\log (cx)$
• D. $y^3=3x^3=\log (cx)$

Answer 1

Answer: (B)

$x{y}^{2}dy-\left(x^3+y^3\right)dx=0$
$\Rightarrow x{y}^{2}dy=\left(x^3+y^3\right)dx$ (changing sides)
$\Rightarrow \frac{dy}{dx}=\frac{x^3+y^3}{x{y}^2}$ ...(i)
Now, Let $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ ..(ii)
replacing value of $\frac{dy}{dx}$ in (i)
$v+x\frac{dv}{dx}=\frac{x^3+y^3}{x{y}^2}$
$=\frac{x^3}{x{y}^2}+\frac{y^3}{y^{2}x}=\frac{x^2}{y^2}+\frac{y}{x}$
$\therefore v+x\frac{dv}{dx}=\frac{x^2}{y^2}+\frac{y}{x}$
$v+x\frac{dv}{dx}=\frac{x^2}{y^2}+v$
$\Rightarrow x\frac{dv}{dx}=\frac{1}{y^2/x^2}=\frac{1}{v^2}$
$\Rightarrow v^{2}dv=\frac{dx}{x}$ ...(iii)
(by cross multiplication)
Now, integrating both side of equation (iii)
we have,
$\int v^{2}dv=\int \frac{dx}{x}$
$\Rightarrow \frac{v^3}{3}=\log x+\log c$
$\Rightarrow \frac{v^3}{3}=\log (xc)$
$[\because \log a+\log b=\log ab]$
$\Rightarrow v^3=3\log (cx)$
$\Rightarrow \frac{y^3}{x^3}=3\log (cx)$
$y^3=3x^3\log (cx)$