Question

Решите дифференциальное уравнение

Answer 1

$y'+yctgx=frac{1}{sinx}\y=uv;y'=u'v+v'u\u'v+v'u+uvctgx=frac{1}{sinx}\u'v+u(v'+vctgx)=frac{1}{sinx}\begin{cases}v'+vctgx=0\u'v=frac{1}{sinx}end{cases}\frac{dv}{dx}+vctgx=0|*frac{dx}{v}\frac{dv}{v}=-ctgxdx\intfrac{dv}{v}=-int ctgxdx\ln|v|=-ln|sinx|\v=frac{1}{sinx}\frac{du}{sinxdx}=frac{1}{sinx}\du=dx\int du=int dx\u=x+C\y=frac{x}{sinx}+frac{C}{sinx}$
$y(frac{pi}{2})=frac{pi}{2}\frac{pi}{2}=frac{pi}{2}+C\C=0\y=frac{x}{sinx}$