Answer to Question #155051 in Differential Equations for almonia

Question

y(x²+1)=cx

Answer 1

$y=\frac{cx}{x^2+1}\\ y'=\frac{cx^2+c-2cx^2}{(x^2+1)^2}\\ y'=\frac{c(1-x^2)}{(x^2+1)^2}\\ \frac{y'}{y}=\frac{c(1-x^2)}{(x^2+1)^2}\times \frac{x^2+1}{cx}\\ y'=\frac{y(1-x^2)}{x^2+1}$