Question

Solve the given D.E:

$(x^2-y^2)dx+2xydy=0$

• A. $(x^2-y^2)=C|x|$
• B. $(x^2+y^2)=C|x|$
• C. $x^2+y^2=C$
• D. none of these

Answer 1

The correct option is B $(x^2+y^2)=C|x|$

Given $(x^2-y^2)dx+2xydy=0$

$(x^2-y^2)dx=-2xydy$

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

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ ......(1)

Since each of the functions $y^2-x^2$ and $2xy$ is a homogeneous function of degree 2, the given differential equation is therefore homogeneous.

Putting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in equation 1, we get,

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

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

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

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

$x\frac{dv}{dx}=-\left(\frac{v^2+1}{2v}\right)$

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

integrating on both sides

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

$log(v^2+1)=-log|x|+C$

$log(v^2+1)+log|x|=logC$

$(v^2+1)|x|=C$

$(x^2+y^2)=C|x|$