Question

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

• A. $a=1,b=0$
• B. $a=3,b=1$
• C. $a=1,b=1$
• D. $a=2,b=2$

Answer 1

Answer: (A)

$a=1,b=0$

$\frac{dy}{dx}=\frac{y}{x^2}$
$\Rightarrow \frac{dy}{y}=\frac{dx}{x^2}=x^{-2}dx$
Integrating both sides, we get
$\ln y=-x^{-1}+c=-\frac{1}{x}+c$
$\Rightarrow y=e^{-1/x+c}=e^c{e}^{-1/x}=a{e}^{-1/x}$
Where $e^c=a=$ any constant.
Comparing it with given solution we get, $b=0,a\in R$