Question

If $\frac{2dy}{dx}=\frac{x}{y}-1$ then

• A. $\left(x+y\right)^{2}\left(2x-y\right)=c$
• B. $\left(x-y\right)^{2}\left(2y-x\right)=c$
• C. $\left(x-y\right)^{2}\left(2x-y\right)=c$
• D. None of these

Answer 1

Answer: (D)

Substitute $y=vx$
$ \Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, given equation becomes $2\left(\frac{xdv}{dx}+v\right)$
$=\frac{1}{v}-1$
$\Rightarrow \frac{xdv}{dx}=\frac{1}{2v}\left(1-v\right)-v$
$=\frac{1-v-2v^{2}}{2v}$
$\Rightarrow \int \frac{dx}{x}=\int \frac{2v\,dv}{\left(1+v\right)\left(1-2v\right)}$
$=-\frac{2}{3} \int \left(\frac{1}{1+v}+\frac{1}{2v-1}\right)dv$
$ \Rightarrow ln\,x+ln\,c_{1}=-\frac{1}{3}ln\left(1+v\right)^{2}\left(2v-1\right)$
$\Rightarrow ln\left(x^{3}\left(1+v\right)^{2}\left(2v-1\right)\right)=ln\,c$
$\Rightarrow x^{3}\left(1+v\right)^{2}\left(2v-1\right)=c$
$\Rightarrow \left(x+y\right)^{2}\left(2y-x\right) = c$.