Question

The solution of $\frac{dy}{dx}=\frac{2xy}{x^2+y^2}$ is

• A. $x^2+y^2=cy$
• B. $y^2-x^2=cy$
• C. $y^2-x^2=cx$
• D. $\frac{y(x-y)}{x(x+y)}$=$xc$
  

Answer 1

The correct option is D

$\frac{y(x-y)}{x(x+y)}$=$xc$

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

$\frac{dy}{dx}=\frac{2\times \frac{y}{x}}{1+(\frac{y}{x}{)}^2}$ ...(i)

Let, $u=\frac{y}{x}$...(ii)

$y=ux$

$\frac{dy}{dx}=x\times \frac{du}{dx}+u$...(iii)

from (i)(ii)&(iii),

$x\times \frac{du}{dx}+u=\frac{2\times u}{1+u^2}$

$x\times \frac{du}{dx}=\frac{2\times u}{1+u^2}-u$

$x\times \frac{du}{dx}=\frac{2\times u-\left[u\right(1+u^2\left)\right]}{1+u^2}$

$x\times \frac{du}{dx}=\frac{u-u^3}{1+u^2}$

$\left(\frac{1+u^2}{u-u^3}\right)du=\frac{dx}{x}$

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

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

simplifying LHS as,

$\frac{1+u^2}{u(1-u)(1+u)}=\frac{A}{u}-\frac{B}{1+u}+\frac{C}{1-u}$

solving for A,B & C by substituting value of $u\in (-1,0,1)$ we obtain, A=1, B&C=-1

$\frac{1+u^2}{u(1-u)(1+u)}=\frac{1}{u}-\frac{1}{1+u}+\frac{1}{1-u}$

$[\frac{1}{u}-\frac{1}{1+u}+\frac{1}{1-u}]du=\frac{dx}{x}$

$\int [\frac{1}{u}+\frac{-1}{1+u}+\frac{1}{1-u}]du=\int \frac{dx}{x}$

$\log u-\log (1+u)+\log (1-u)=\log x+\log c$

$\log \frac{u(1-u)}{1+u}$=$\log \left(xc\right)$

$\frac{u(1-u)}{u+1}$=$xc$

substituting the value of u

$\frac{y(x-y)}{x(x+y)}$=$xc$