Question

y'tgx-y=0 дифференциальное уравнение

Answer 1

$y'\, tgx-y=0\\\\y'=\frac{y}{tgx}\\\\\int \frac{dy}{y}=\int \frac{dx}{tgx}\\\\ln|y|=\int ctgx\, dx\\\\ln|y|=ln|sinx|+lnC\\\\y=C\, sinx\\\\\\\star \int ctgx\, dx=\int \frac{cosx\, dx}{sinx}=\int \frac{d(sinx)}{sinx}=\int \frac{du}{u}=ln|u|+C=ln|sinx|+C$