Question

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation, $y(1 + xy) dx = x \,dy$, then $f \left( - \frac{1}{2} \right)$ is equal to :

• A. $ - \frac{2}{5}$
• B. $ - \frac{4}{5}$
• C. $ \frac{2}{5}$
• D. $ \frac{4}{5}$

Answer 1

The Correct Option is D

$\frac{ y }{ x }(1+ xy )=\frac{ dy }{ dx }$
$y = vx$
$ \Rightarrow \frac{ y }{ x }= v$
$\frac{ dy }{ dx }= v + x \frac{ dv }{ dx }$
$v\left(1+v x^{2}\right)=v+x \frac{d v}{d x}$
$v ^{2} x ^{2}= x \frac{ dv }{ dx }$
$v ^{2} x =\frac{ dv }{ dx }$
$\int xdx =\int \frac{1}{ v ^{2}} dv$
$\frac{x^{2}}{2}=-\frac{1}{v}+c$
$\frac{x^{2}}{2}=-\frac{x}{y}+c$
Put $(1,-1)$
$\frac{1}{2}=\frac{1}{1}+ c $
$\Rightarrow c =\frac{-1}{2}$
$\frac{x^{2}}{2}=-\frac{x}{y}-\frac{1}{2}$
We have to find $f\left(-\frac{1}{2}\right)$
Put $x =-\frac{1}{2}$
$\frac{\left(-\frac{1}{2}\right)^{2}}{2}=\frac{-\left(-\frac{1}{2}\right)}{ y }-\frac{1}{2}$
$\frac{1}{8}=\frac{1}{2 y }-\frac{1}{2}$
$y =\frac{4}{5}$