Question

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

• 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)

$2x^{2}dy=\left(2xy+y^2\right)dx$
$\Rightarrow \frac{dy}{dx}=\frac{2xy+y^2}{2x^2}\{$ Homogeneous D.E. $\}$
$\left\{\begin{array}{c}lety=xt\\ \Rightarrow \frac{dy}{dt}=t+x\frac{dt}{dx}\end{array}\right\}$
$\Rightarrow t+x\frac{dt}{dx}=\frac{2x^{2}t+x^2{t}^2}{2x^2}$
$\Rightarrow t+x\frac{dt}{dx}=t+\frac{t^2}{2}$
$\Rightarrow x\frac{dt}{dx}=\frac{t^2}{2}$
$\Rightarrow 2\int \frac{dt}{t^2}=\int \frac{dx}{x}$
$\Rightarrow 2\left(-\frac{1}{t}\right)=\ell n(x)+C\{$ Put $t=\frac{y}{x}\}$
$\Rightarrow -\frac{2x}{y}=\ell nx+C\left\{\begin{array}{c}Putx1=\&y=2\\ \text{then we get}C=-1\end{array}\right\}$
$\Rightarrow \frac{-2x}{y}=\ell n(x)-1$
$\Rightarrow y=\frac{2x}{1-\ell nx}$
$\Rightarrow f(x)=\frac{2x}{1-{\log }_{e}x}$
so, $\sqrt{f\left(\frac{1}{2}\right)=\frac{1}{1+{\log }_{e}2}}$