Y'+y=e^(x/2) sqrt y дифференциал

Дифференциальное уравнение Бернулли:
$y`+y=e^\frac{x}{2}y^\frac{1}{2}\\\frac{y`}{y^\frac{1}{2}}+y^\frac{1}{2}=e^\frac{x}{2}\\\\z=y^\frac{1}{2}=\ \textgreater \ z`=\frac{y`}{2\sqrt y}\\\frac{y`}{\sqrt y}=2z`\\2z`+z=e^\frac{x}{2}\\z=uv\ ;z`=u`v+v`u\\2u`v+2v`u+uv=e^\frac{x}{2}\\2u`v+u(2v`+v)=e^\frac{x}{2}\\\begin{cases}2v`+v=0\\2u`v=e^\frac{x}{2}\end{cases}$
$2\frac{dv}{dx}=-v\\\frac{dv}{v}=-\frac{dx}{2}\\\int\frac{dv}{v}=-\frac{1}{2}\int dx\\ln|v|=-\frac{x}{2}\\v=e^{-\frac{x}{2}}=\frac{1}{e^\frac{x}{2}}\\\frac{2u`}{e^\frac{x}{2}}=e^{\frac{x}{2}}\\u`=\frac{e^x}{2}\\u=\frac{1}{2}\int e^xdx=\frac{e^x}{2}+C\\z=uv=(\frac{e^x}{2}+C)*\frac{1}{e^\frac{x}{2}}=\frac{e^\frac{x}{2}}{2}+\frac{C}{e^\frac{x}{2}}\\z=\sqrt y=\ \textgreater \ y=z^2\\y=(\frac{e^\frac{x}{2}}{2}+\frac{C}{e^\frac{x}{2}})^2=\frac{e^x}{4}+C+\frac{C^2}{e^x}\\OTBET:y=\frac{e^x}{4}+\frac{C^2}{e^x}+C;C-const;y=0$