Question

Find the integral curve of the differential equation $x(1-x/ny).\frac{dy}{dx}+y=0$, which passes through $(1,\frac{1}{e})$.

Answer 1

$x(1-\frac{x}{ny}).\frac{dy}{dx}+y=0$

$\Rightarrow \frac{dy}{dx}=\frac{y/x}{x/ny-1}=\frac{y/x}{\frac{1}{n.\left(y/x\right)}-1}$

$y=vx$

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

$\Rightarrow v+x\frac{dv}{dx}=\frac{v}{\frac{1}{nv}-1}$

$\Rightarrow x\frac{dv}{dx}=v[\frac{1}{\frac{1}{nv}-1}-1]$

$\Rightarrow x\frac{dv}{dx}=v\left[\frac{2nv-1}{1-nv}\right]$

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

$\Rightarrow \frac{nv-1}{v.(2nv-1)}dv=\frac{-dx}{x}$

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

$\Rightarrow \int \frac{1}{v}dv-\int \frac{n}{(2nv-1)}dv=\int \frac{-dx}{x}$

$\Rightarrow \ln v-\frac{ln(2nv-1)}{2n}=-lnx+c$

$\Rightarrow \ln \left(vn\right)-\frac{\ln (2nv-1)}{2n}=c$

$\Rightarrow \ln \left(y\right)-\frac{\ln (2ny/n-1)}{2n}=c$

At$x=1,y=1/e$

Solving further we get ; $c=\frac{\ln (2n/e-1)}{2n}$

$=\frac{y^{2n}}{2n\frac{y}{n}-1}=\frac{2n}{e}-1$