Question

Решить дифференциальное уравнение xy'-y=x^2cosx срочно

Answer 1

$xy'-y=x^2cosx\\\\y'-\frac{y}{x}=xcosx\\\\y=uv\\\\u'v+uv'-\frac{uv}{x}=xcosx\\\\u'v+u(v'-\frac{v}{x})=xcosx\\\\1)\; v'-\frac{v}{x}=0\\\\\int \frac{dv}{v}=\int \frac{dx}{x}\\\\ln|v|=ln|x|\\\\v=x\\\\2)\;$

$2)\; u'x=xcosx\\\\\int du=\int cosxdx\\\\u=sinx+C\\\\3)\; y=x(sinx+C)$