Question

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

• A. $x^3-y^3=c{x}^2$
• B. $x^3=y^3$
• C. $x^3-y^3=cx$
• D. $x^3+y^3=c{x}^2$

Answer 1

Answer: (D)

$x^3+y^3=c{x}^2$

$\Rightarrow \frac{dy}{dx}=\frac{2y^3-x^3}{3x{y}^2}$

put $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}=\frac{2v^3-1}{3v^2}$

$\Rightarrow x\frac{dv}{dx}=\frac{2v^3-1-3v^3}{3v^2}$

$\Rightarrow \int -\left(\frac{3v^2}{1+v^3}\right)dv=\int \frac{dx}{x}-\log c$

$\Rightarrow \log c=\log x+\log (1+v^3)=\log (1+v^3)x$

$\Rightarrow x(1+v^3)=c$

$\Rightarrow x(1+\frac{y^3}{x^3})=c$

$\Rightarrow x^3+y^3=c{x}^2$