Question

Solution of the differential equation $ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$ is

• A. $\log \left(\frac{y}{x}\right)+1=Cy$
• B. $\log \left(\frac{y}{x}\right)-1=Cy$
• C. $\log \left(\frac{y}{x}\right)-1=Cx$
• D. $noneofthese$

Answer 1

The correct option is B $\log \left(\frac{y}{x}\right)-1=Cy$

$ydx+d\log \left(y/x\right)dy-2xdy=0$

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

Putting $F(x,y)=\frac{dy}{dx}$ and finding $F(\lambda x,\lambda y)$

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

$F(\lambda x,\lambda y)=\frac{\frac{\lambda y}{\lambda x}}{2-\log \left(\frac{\lambda y}{\lambda x}\right)}=\frac{y/x}{2-\log \left(y/x\right)}={\lambda }^o\left[F\right(x,y\left)\right]$

This $,F(x,y)$ is a homogeneous eq function of order $o$ therefore $\frac{dy}{dx}$ is homogeneous differential equation solving $\frac{dy}{dx}$ by putting $y=vx$

$y=vx$

$\frac{dy}{dx}=\frac{xdv}{dx}+\frac{vdx}{dx}\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$

Putting value of $\frac{dy}{dx}$ and $y=vx$ in $\left(i\right)$

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

$v+\frac{xdv}{dx}=\frac{vx/x}{2-\log \left(\frac{vx}{x}\right)}=\frac{v}{2-\log v}$

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

$\int \frac{r-\log v}{v\log v}dv=\int \frac{dx}{x}\Rightarrow \int \frac{2-\log v}{-v(1-\log v)}dx=\log x+\log c$

By solving

$\log (\log y/x-1)=\log xcy/x$

$\log y/x-1=cy$

$cy=\log |y/x|-1$