Question

y'+y/x=sin x
Помогите

Answer 1

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

$2); ; u'v=sinx\\frac{du}{dx}=frac{sinx}{v}=frac{sinx}{frac{1}{x}}=xsinx\\int du=int x, sinxcdot dx\\int x, sinxcdot dx=[u=x,, du=dx,; dv=sinx, dx,; v=-cosx]=\\=-x, cosx-int (-cosx)dx=-x, cosx+sinx+C\\3); ; y=uv=frac{1}{x}(-xcosx+sinx+C)=-cosx+frac{sinx}{x}+Cx$