Question

Solution of $y^{2}dx=(xy-x^2)dy$, given that $y=1$ when $x=1$, is:

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

Answer 1

The correct option is A $(1+\log y)x=y$

$\Rightarrow \frac{dy}{dx}=\frac{y^2}{xy-x^2}$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2}{v-1}$

$x\frac{dv}{dx}=\frac{v^2-v^2+v}{v-1}$

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

$\Rightarrow \int (1-\frac{1}{v})dv=\log cx$
$\Rightarrow v-\log v=\log cx$
$\Rightarrow v-\log y+\log x=\log x+\log c$
$\Rightarrow y=x\log c+x\log y$

Substitute $(x,y)=(1,1)$
$\Rightarrow 1=x\log c+0\Rightarrow \log c=1$
$\Rightarrow y=x(1+\log y)$

Hence, option 'A' is correct.