Question

$y=a{e}^{-1/x}+b$ is a solution of the differential equation $\frac{dy}{dx}=\frac{y}{x^2}$ when

• A. $a=2,b=1$
• B. $aϵR,b=0$
• C. $a=1,b=1$
• D. none of these

Answer 1

Answer: (B)

$aϵR,b=0$

$\frac{dy}{dx}=\frac{y}{x^2}$

$\therefore \frac{dy}{dx}=\frac{dx}{x^2}$

$\therefore \int \frac{dy}{y}=\int \frac{dx}{x^2}$

$\therefore \ln y+4=\frac{-1}{x}+C_2$

$\therefore \ln y=\frac{-1}{x}+C$

$\therefore y=e^{-1/x+C}$

$y=k{e}^{-1/x}[k=e^C]$

$\therefore$For $y=a{e}^{-1/x}+b$ to be a solution, $aϵR$ & $b=0$

Thus, the correct option is $aϵR,b=0$