Question

The solution of the differential equation $(x^2-y^2)dx+2xydy=0$, is.

• A. $x^2-y^2=Cx$
• B. $x^2+y^2=Cy$
• C. $x^2+y^2=Cx$
• D. $x^2-y^2=Cy$

Answer 1

Answer: (C)

$x^2+y^2=Cx$

Given D.E. is $(x^2-y^2)dx+2xydy=0$

$⟹\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ ....... $\left(i\right)$

$\therefore$ It is homogeneous

Take $y=vx$

$⟹\frac{dy}{dx}=x\frac{dv}{dx}+v\frac{dx}{dx}$

$⟹\frac{dy}{dx}=x\frac{dv}{dx}+v$

Substituting $y$ and $\frac{dy}{dx}$ in $\left(i\right)$ we get

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

$⟹x\frac{dv}{dx}+v=\frac{v^2-1}{2v}$

$⟹x\frac{dv}{dx}=\frac{-v^2-1}{2v}$

$⟹-\frac{1}{x}dx=\frac{2v}{v^2+1}dv$

Taking integration both sides we get

$⟹\int -\frac{1}{x}dx=\int \frac{2v}{v^2+1}dv$

$⟹-\ln |x|=\ln |v^2+1|+C$

$⟹\ln |v^2+1|+\ln |x|=C$

$⟹\ln \left(\right(v^2+1\left)x\right)=C$

$⟹\ln \left((\frac{y^2}{x^2}+1)x\right)=C$

$⟹\ln \left(\frac{x^2+y^2}{x}\right)=C$

$⟹\frac{x^2+y^2}{x}=C$

$⟹x^2+y^2=Cx$