Question

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

• A. $x^2+y^2=cx$
• B. $x^2-y^2+cx=0$
• C. $x^2+2xy=y^2+cx$
• D. $x^2+y^2=2xy+c{x}^2$

Answer 1

Answer: (A)

$x^2+y^2=cx$

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

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

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

putting $y=vx$

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

Now, $v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2x\times vx}$

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

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

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

putting $v^2+1=t$

$2vdv=dt$

$\Rightarrow -{\int }^{}\frac{dt}{t}={\int }^{}\frac{dx}{x}$

$\Rightarrow -\mathrm{logt}=\log x+c_1$

$\Rightarrow \frac{1}{t}=x{c}_1$

$\Rightarrow \frac{1}{v^2+1}=x{c}_1$

$\Rightarrow \frac{1}{\frac{y^2}{x^2}+1}=x{c}_1$

$\Rightarrow \frac{x^2}{x^2+y^2}=x{c}_1$

$\Rightarrow x^2+y^2=\frac{x}{c_1}$

$\Rightarrow x^2+y^2=xc$