Answer
$y=e^{\frac{-1}{x}+1}$
Work Step by Step
We are given that $x^2 \dfrac{dy}{dx}=y$
We will separate the variables to obtain:
$\dfrac{\ dy}{y}=\dfrac{\ dx}{x^2}$
Integrate to obtain:
$\int \dfrac{\ dy}{y}=\int \dfrac{\ dx}{x^2}$
This implies that $\ln |y|=\dfrac{-1}{x}+C$
After applying the initial conditions, $y=1$ when $x=1$, we get $C=1$
Therefore, we have: $\ln |y|=\dfrac{-1}{x}+1 \implies y=e^{\frac{-1}{x}+1}$
