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 $2\left(\frac{xdv}{dx}+v\right)$
$=\frac{1}{v}-1$
$\Rightarrow \frac{xdv}{dx}=\frac{1}{2v}\left(1-v\right)-v$
$=\frac{1-v-2v^2}{2v}$
$\Rightarrow \int \frac{dx}{x}=\int \frac{2vdv}{\left(1+v\right)\left(1-2v\right)}$
$=-\frac{2}{3}\int \left(\frac{1}{1+v}+\frac{1}{2v-1}\right)dv$
$\Rightarrow lnx+ln{c}_1=-\frac{1}{3}ln{\left(1+v\right)}^2\left(2v-1\right)$
$\Rightarrow ln\left(x^3{\left(1+v\right)}^2\left(2v-1\right)\right)=lnc$
$\Rightarrow x^3{\left(1+v\right)}^2\left(2v-1\right)=c$
$\Rightarrow {\left(x+y\right)}^2\left(2y-x\right)=c$.