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

$\frac{1}{(1+{\log }_{e}2)}$

Explanation for the correct option:

Step 1. Find the value of $f\left(\frac{1}{2}\right)$:

Given, $2x^{2}dy=(2xy+y^2)dx$ …..(1)

$\Rightarrow$ $\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}$ {Homogeneous D.E}

Let $y=xt$

$\Rightarrow$$\frac{dy}{dx}=t+x\frac{dt}{dx}$

Step 2. Put the value of $\frac{dy}{dx}$ in equation (1), we get

$t+x\frac{dt}{dx}=\frac{2x^{2}t+x^2{t}^2}{2x^2}$

$\Rightarrow$$t+x\frac{dt}{dx}=t+\frac{t^2}{2}$

$\Rightarrow$ $x\frac{dt}{dx}=\frac{t^2}{2}$

$\Rightarrow$ $2\int \frac{dt}{t^2}=\int \frac{dx}{x}$

$\Rightarrow$ $2\left(-\frac{1}{t}\right)=\ln x+C$

Put $t=\frac{y}{x}$

$\Rightarrow$$-\frac{2x}{y}=\ln x+C$

Step 3. At $x=1$ and $y=2$, we get

$-\frac{2\times 1}{2}=\ln 1+C$

$\Rightarrow$ $C=-1$

$\Rightarrow$$-\frac{2x}{y}=\ln x-1$

$\Rightarrow$ $y=\frac{2x}{1-\ln x}$

$\Rightarrow$ $f\left(x\right)=\frac{2x}{1-{\log }_{e}x}$

$\mathbf{\therefore }\mathit{f}\left(\frac{1}{2}\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{e}}\mathbf{2}}$

Hence, Option ‘C’ is Correct.