Question

Solve the differential equation: $2\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x^2}.$

• A. $2x=e^y(2c{x}^2-1)$
• B. $2x=-e^y(2c{x}^2+1)$
• C. $2x=-e^y(2c{x}^2-1)$
• D. None of these.

Answer 1

Answer: (C)

$2x=-e^y(2c{x}^2-1)$

$2\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x^2}\Rightarrow e^{-y}\frac{dy}{dx}+\frac{e^{-y}}{2x}=\frac{1}{2x^2}$
Put $-e^{-y}=v\Rightarrow e^{-y}dy=dv$
$\therefore \frac{dv}{dx}-\frac{v}{x}=\frac{1}{x^2}$ ...(1)
Here $P=-\frac{1}{x}\Rightarrow \int Pdx=-\int \frac{1}{x}dx=-\log x=\log \frac{1}{x}$
$\therefore I.F.=e^{\log \frac{1}{x}}=\frac{1}{x}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{x}\frac{dv}{dx}-\frac{v}{x^2}=\frac{1}{x^3}$
Integrating both sides
$\frac{v}{x}=\int \frac{1}{x^3}dx+c=-\frac{1}{{2x}^2}+c$

$\Rightarrow 2x=-e^y(2c{x}^2-1)$