Question

Let $y\left(x\right)$ be the solution of the differential equation $2x^{2}dy+(e^y-2x)dx=0,x>0.$ If $y\left(e\right)=1$, then $y\left(1\right)$ is equal to

• A. $0$
• B. $2$
• C. ${\log }_e\left(2e\right)$
• D. ${\log }_e\left(2\right)$

Answer 1

The correct option is D ${\log }_e\left(2\right)$

$\frac{dy}{dx}=\frac{2x-e^y}{2x^2}$
$\Rightarrow -e^{-y}\frac{dy}{dx}+\frac{e^{-y}}{x}=\frac{1}{2x^2}\cdots \left(i\right)$
Let $e^{-y}=t$
$-e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$
Putting this in equation $\left(i\right)$
$\Rightarrow \frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$
$\Rightarrow \frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$
This is a first order linear differential equation
$I.F.=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$
$\Rightarrow t\cdot x=\int \frac{1}{2x^2}\times xdx+C$
$\Rightarrow t\cdot x=\int \frac{1}{2x}dx+C$
$\Rightarrow x{e}^{-y}=\frac{1}{2}\ln x+C$
We know $y\left(e\right)=1$
$\Rightarrow e\cdot e^{-1}=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}$
$\Rightarrow x{e}^{-y}=\frac{1}{2}\ln x+\frac{1}{2}$
Now put $x=1,$
$\Rightarrow 1\times e^{-y}=\frac{1}{2}\ln 1+\frac{1}{2}$
$\Rightarrow e^{-y}=0+\frac{1}{2}$
$\Rightarrow -y\ln e=-\ln 2$
$\Rightarrow y=\ln 2$