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. $y^3=3x^{3}log\left(cx\right)$
• C. $y^3=3x^3+log\left(cx\right)$
• D. $y^3+3x^3+log\left(cx\right)$

Answer 1

The correct option is B $y^3=3x^{3}log\left(cx\right)$

$x{y}^{2}dy-(x^3+y^3)dx=0$

$\Rightarrow x{y}^{2}dy=(x^3+y^3)dx$ ....(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,

$\int v^{2}dv=\int \frac{dx}{x}$

$\Rightarrow \frac{v^3}{3}=\log x+\log c$

$\to \frac{v^3}{3}=\log \left(xc\right)$ $[\because \log a+\log b=\log ab]$

$\Rightarrow v^3=3\log \left(cx\right)$

$\Rightarrow \frac{y^3}{x^3}=3\log \left(cx\right)$

$y^3=3x^3\log \left(cx\right)$

So, the solution is, $y^3=3x^3\log \left(cx\right)$

Hence, the option is B