Question

Solve the differential equation $(x^2-y^2)dx-xydy=0$.

• A. $y^2(x^2-2y^2)=k$
• B. $x^2(x^2-2y^2)=k$
• C. $y^2(2x^2-y^2)=k$
• D. $x^2(2x^2-y^2)=k$

Answer 1

Answer: (B)

$x^2(x^2-2y^2)=k$

$(x^2-y^2)dx-xydy=0\Rightarrow 2y\frac{dy}{dx}+\frac{2y^2}{x}=2x$
Substitute $v=y^2\Rightarrow \frac{dv}{dx}=2y\frac{dy}{dx}$
$\therefore \frac{dv}{dx}+\frac{2v}{x}=2x$ ...(1)
Here $P=\frac{2}{x}\Rightarrow \int Pdx=\int \frac{2}{x}dx=2\log x=\log x^2$
$\therefore I.F.=e^{\log x^2}=x^2$
Multiplying (1) by $I.F.$ we get
$x^2\frac{dv}{dx}+2vx=2x^3$
Integrating both sides we get
$x^{2}v=2\int x^{3}dx+c=\frac{x^4}{2}+c\Rightarrow x^2(x^2-2y^2)=C=k$