Question

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

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

Answer 1

The correct option is C $\frac{1}{1+{\log }_{e}2}$

$2\frac{dy}{dx}=2\frac{y}{x}+{\left(\frac{y}{x}\right)}^2$
This is an Homogeneous Differential Equation.

Applying $y=vx$
$2(v+x\frac{dv}{dx})=2v+v^2$
$\Rightarrow 2\frac{dv}{v^2}=\frac{dx}{x}$
$\Rightarrow -\frac{2}{v}=\ln x+c$
$\Rightarrow -\frac{2x}{y}=\ln x+c\cdots \left(1\right)$

Given $f\left(x\right)$ passes through $(1,2)$
From $\left(1\right),$
$c=-1$

Thus y=$f\left(x\right)=-\frac{2x}{(logx-1)}$

For $x=\frac{1}{2}$ we have $f\left(\frac{1}{2}\right)=\frac{1}{1+\ln 2}$