Question

Solve the differential equation: $(x^3-2y^3)dx+3x{y}^{2}dy=0$

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

Answer 1

Answer: (A), (B)

$x^3+y^3=k{x}^2$
$x^3+y^3=-k{x}^2$

$(x^3-2y^3)dx+3x{y}^{2}dy=0\Rightarrow 3y^2\frac{dy}{dx}-\frac{2y^3}{x}=-x^3$
Substitute $v=y^3\Rightarrow dv=3y^{2}dy$
$\therefore \frac{dv}{dx}-\frac{2v}{x}=-x^2$ ...(1)
Here $P=-\frac{2}{x}\Rightarrow \int Pdx=-\int \frac{2}{x}dx=-2\log x=\log \frac{1}{x^2}$
$\therefore I.F.=e^{\log \frac{1}{x^2}}=\frac{1}{x^2}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{x^2}\frac{dv}{dx}-\frac{2v}{x^3}=-1$
Integrating both sides w.r.t $x$ we get
$\frac{v}{x^2}=-\int dx+k=-x+k\Rightarrow x^3+y^3=\pm k{x}^2$