Answer to Question #241160 in Differential Equations for Jojo

Question

(y ^ 2 * sin x) * d * x + (1 / x - y / x) * d * y = 0

Answer 1

$y^2\sin xdx+\frac{1-y}{x}dy=0\\ \frac{1-y}{y^2}dy=-x\sin xdx\\ \text{Integrate both sides}\\ \int \frac{1}{y^2}-\frac{1}{y}dy=-\int x\sin x dx\\ -\frac{1}{y}-\ln y=-[-x\cos x-\int -\cos x dx]\\ -\frac{1}{y}-\ln y=x\cos x-\sin x +C\\ e^{\frac{1}{y}+\ln y}=e^{-[x\cos x-\sin x +C]}\\ ye^{\frac{1}{y}}=e^{-[x\cos x-\sin x +C]}$