Question

Solve: $\frac{dy}{dx}=\frac{x(2y-x)}{x(2y+x)}$, if $y=1$ when $x=1$

Answer 1

Given,

$\frac{dy}{dx}=\frac{x(2y-x)}{x(2y+x)}$

$\frac{dy}{dx}=\frac{(2y-x)}{(2y+x)}$

substitute $y=vx\to \frac{dy}{dx}=x\frac{dv}{dx}+v$

$x\frac{dv}{dx}+v=\frac{(2vx-x)}{(2vx+x)}$

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

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

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

integrating on both sides, we get,

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

$-\frac{1}{2}{v}^2+v-\log (2v+1)+\frac{5}{8}=\log x+c$

$-\frac{1}{2}{\left(\frac{y}{x}\right)}^2+\frac{y}{x}-\log (2\frac{y}{x}+1)+\frac{5}{8}=\log x+c$

put $y=1=x$, we get,

$-\frac{1}{2}{\left(\frac{1}{1}\right)}^2+\frac{1}{1}-\log (2\frac{1}{1}+1)+\frac{5}{8}=\log 1+c$

$-\frac{1}{2}+1+\frac{5}{8}=c$

$\therefore c=\frac{9}{8}$

$-\frac{1}{2}{\left(\frac{y}{x}\right)}^2+\frac{y}{x}-\log (2\frac{y}{x}+1)+\frac{5}{8}=\log x+\frac{9}{8}$

$\therefore -\frac{1}{2}{\left(\frac{y}{x}\right)}^2+\frac{y}{x}-\log (2\frac{y}{x}+1)=\log x+\frac{1}{2}$