Question

Show that the differential equation
$ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$
is homogenous and solve it.

Answer 1

Given differential equation:
$ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$

$dy[2x-x\log \left(\frac{y}{x}\right)]=ydx$

$\frac{dy}{dx}=\frac{y}{x(2-\log \left(\frac{y}{x}\right))}$

$\frac{dy}{dx}=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)}$

If $F(\lambda x,\lambda y)=\lambda F(x,y)$, then the differential equation is homogeneous.

Now,

Substituting $F(x,y)=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)}$,

$F(\lambda x,\lambda y)=\frac{\frac{\lambda y}{\lambda x}}{2-\log \left(\frac{\lambda y}{\lambda x}\right)}$
$F(\lambda x,\lambda y)=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)}=F(x,y)$

Hence, the differential equation is
homogeneous.

Given differential equation :
$ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$
$\frac{dy}{dx}=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)}$ ...$\left(i\right)$
Let $y=vx$

Differentiating both sides $w.r.t{.}^{\prime }{x}^{\prime }$, we get
$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$
Using product rule : ${\left(uv\right)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

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

Substituting the value of $\frac{dy}{dx}$ and $y=vx$ in
$\left(i\right)$, we get
$v+\frac{xdv}{dx}=\frac{\frac{vx}{x}}{2-\log \left(\frac{vx}{x}\right)}$
$v+\frac{xdv}{dx}=\frac{v}{2-\log v}$

$\frac{xdv}{dx}=\frac{v}{2-\log v}-v$
$\frac{xdv}{dx}=\frac{v-2v+v\log v}{2-\log v}$
$\frac{xdv}{dx}=\frac{v\log v-v}{2-\log v}$
$\frac{2-\log v}{v\log v-v}dv=\frac{dx}{x}$

Integrating both sides, we get
$\int \frac{2-\log v}{v\log v-v}dv=\int \frac{dx}{x}$
$\int \frac{2-\log v}{-v(1-\log v)}dv=\log |x|+\log c$
$\int \frac{1+1-\log v}{-v(1-\log v)}dv=\log |x|+\log c$
$\int \frac{dv}{(-v)(1-\log v)}-\int \frac{1}{v}dv$
$=\log |x|+\log c$
$\int \frac{dv}{v(\log v-1)}-\log |v|=\log |x|+\log c$
...$\left(ii\right)$

Substituting $t=\log v-1$,
$dt=\frac{1}{v}dv$
Substituting in $\left(ii\right)$,
$\int \frac{dt}{t}-\log |v|=\log |x|+\log c$
$\log |t|-\log |v|=\log |x|+\log c$
Substituting value of $t$, we get

$\log (\log v-1)-\log v=\log x+\log c$
$\log (\log v-1)=\log x+\log c+\log v$
$\log (\log v-1)=\log cxv$
$(\because \log a+\log b=\log ab)$
$\log v-1=cxv$

Substituting $v=\frac{y}{x}$, we get
$\log \frac{y}{x}-1=cy$

Hence, the required general solution is
$\log \frac{y}{x}-1=cy$