Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 21

$$ y =e^{-\sqrt{1-x^2}+c}.$$

We have $$\sqrt{1-x^2}y'=xy\Longrightarrow \frac{dy}{y}=\frac{xdx}{\sqrt{1-x^2}}\\ \Longrightarrow\int \frac{dy}{y}=\int\frac{xdx}{\sqrt{1-x^2}}+c\\ \Longrightarrow \ln y=-\frac{1}{2} 2\sqrt{1-x^2}+c \Longrightarrow y =e^{-\sqrt{1-x^2}+c}.$$ Hence, the general solution is $$ y =e^{-\sqrt{1-x^2}+c}.$$