Question

Y'-yctgx=sin2x*cosx y(pi/2)=0 Задача Коши.

Answer 1

$y'-yctgx=sin2xcosx\y=uv;y'=u'v+v'u\u'v+v'u-uvctgx=sin2xcosx\u'v+u(v'-vctgx)=sin2xcosx\begin{cases}v'-vctgx=0\u'v=sin2xcosxend{cases}\frac{dv}{dx}-vctgx=0|frac{dx}{v}\frac{dv}{v}=ctgxdx\intfrac{dv}{v}=intfrac{d(sinx)}{sinx}\ln|v|=ln|sinx|\v=sinx\frac{du}{dx}sinx=sin2xcosx\frac{du}{dx}=2cos^2x\du=2cos^2xdx\int du=int (1+cos2x)dx\u=x+frac{1}{2}sin2x+C\y=xsinx+frac{1}{2}sin2xsinx+Csinx\y(frac{pi}{2})=0\0=frac{pi}{2}+C\C=-frac{pi}{2}$
$y=xsinx+frac{1}{2}sin2xsinx-frac{pi}{2}sinx$