Question

The solution of $\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x^2}$ is

• A. $2x=\left(1+C{x}^2\right)e^y$
• B. $x=\left(1+C{x}^2\right)e^y$
• C. $2x^2=\left(1+C{x}^2\right)e^{-y}$
• D. $x^2=\left(1+C{x}^2\right)e^{-y}$

Answer 1

Answer: (A)

We have, $\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x^2}$
$\Rightarrow e^{-y}\frac{dy}{dx}+\frac{1}{x}{e}^{-y}=\frac{1}{x^2}$
Put $e^{-y}=v$
$\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dv}{dx}$
$\therefore \frac{dv}{dx}-\frac{1}{x}v=\frac{-1}{x^2}$
Here, above equation is a linear differential equation in $v$.
$\therefore IF=e^{\int -\frac{1}{x}dx}=e^{-\log x}=\frac{1}{X}$
Hence, the required solution is
$v\cdot \left(\frac{1}{x}\right)=\int \frac{-1}{x^2}\cdot \frac{1}{x}dx+C_1$
$\Rightarrow \frac{V}{X}=\frac{1}{2x^2}+C_1$
$\Rightarrow \frac{V}{X}=\frac{1+2x^2{C}_1}{2x^2}$
$\Rightarrow 2xv=1+C{x}^2[\because 2C_1=C]$
$\Rightarrow 2x{e}^{-y}=1+C{x}^2[\because v=e^{-y}]$
$\Rightarrow 2x=(1+C{x}^2)e^y$