Question

If a curve $y=f(x)$, passing through the point $(1,2),$ is the solution of the differential equation. $2 x ^{2} dy =\left(2 xy + y ^{2}\right) dx ,$ then $f \left(\frac{1}{2}\right)$ is equal to.
So, $f(\frac{1}{2}) = \frac{1}{1 + log_e \,2}$

• A. $\frac{1}{1-\log _{ e } 2}$
• B. $\frac{1}{1+\log _{e} 2}$
• C. $\frac{-1}{1+\log _{ e } 2}$
• D. $1+\log _{ e } 2$

Answer 1

Answer: (B)

$2 x^{2} d y=\left(2 x y+y^{2}\right) d x$
$\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\{$ Homogeneous D.E. $\}$
$\begin{Bmatrix}let y=xt\\ \Rightarrow \frac{dy}{dt}=t+x \frac{dt}{dx}\end{Bmatrix}$
$\Rightarrow t + x \frac{ dt }{ dx }=\frac{2 x ^{2} t + x ^{2} t ^{2}}{2 x ^{2}}$
$\Rightarrow t + x \frac{ dt }{ dx }= t +\frac{ t ^{2}}{2}$
$\Rightarrow x \frac{ dt }{ dx }=\frac{ t ^{2}}{2}$
$\Rightarrow 2 \int \frac{ dt }{ t ^{2}}=\int \frac{ d x }{ x }$
$\Rightarrow 2\left(-\frac{1}{t}\right)=\ell n(x)+C\left\{\right.$ Put $\left.t=\frac{y}{x}\right\}$
$\Rightarrow -\frac{2 x}{y}=\ell n x+C \begin{Bmatrix}Put x 1=\&y=2\\ \text{then we get} C=-1\end{Bmatrix}$
$\Rightarrow \frac{-2 x}{y}=\ell n(x)-1$
$\Rightarrow y =\frac{2 x }{1-\ell n x }$
$\Rightarrow f(x)=\frac{2 x}{1-\log _{e} x}$
so, $\sqrt{f\left(\frac{1}{2}\right)=\frac{1}{1+\log _{e} 2}}$