Question

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

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

Answer 1

Answer: (A)

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