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

• Correct answer
  A

  $2 x=\left(1+C x^2\right) e^y$
• B

  $x =\left(1+C x^2\right) e^y$
• C

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

  $x^2=\left(1+C x^2\right) e^{-y}$

Solution

Given differential equation, $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$ Dividing the eqn. by $e^y$, we get
Let $e^{-y}=v$ $\Rightarrow \quad e^{-y} \frac{d y}{d x}=\frac{d v}{d x}$ From eqn. (i), we get $\therefore \quad \frac{d v}{d x}-\frac{1}{x} v=\frac{-1}{x^2}$ Integrating Factor, $\mathrm{IF}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log \frac{1}{x}}=\frac{1}{x}$ Now, the required solution is $\begin{aligned} & v(\mathrm{IF})=\int-\frac{1}{x^2}(\mathrm{IF}) d x+C_1 \\ & v \cdot\left(\frac{1}{x}\right)=\int \frac{-1}{x^2} \cdot \frac{1}{x} d x+C_1 \\ & \frac{v}{x}=-\int x^{-3} d x+C_1 \\ & \frac{v}{x}=\frac{1}{2 x^2}+C_1 \Rightarrow \frac{v}{x}=\frac{1+2 x^2 C_1}{2 x^2} \\ & 2 x v=1+C x^2 \quad\left[\because 2 C_1=C\right] \\ & 2 x e^{-y}=1+C x^2 \quad\left[\because v=e^{-y}\right] \\ & 2 x=\left(1+C x^2\right) e^y \quad\left[\because{ }^y\right. \end{aligned}$

Hence, the correct answer is option: 'A'.