What is the solution for 2y^'=e^{x/2}+y ?

The solution for 2y^'=e^{x/2}+y is y= x/(2e^{-x/2)}+c_{1}e^{x/2}

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2y^'=e^{x/2}+y

2 y^{^{'}} = e^{\frac{x}{2}} + y

y = \frac{x}{2 e ^{- \frac{x}{2}}} + c_{1} e^{\frac{x}{2}}

Solution steps

2 y^{'} (x) = e^{\frac{x}{2}} + y

Solve linear ODE:

y = \frac{x}{2 e ^{- \frac{x}{2}}} + c_{1} e^{\frac{x}{2}}

y = \frac{x}{2 e ^{- \frac{x}{2}}} + c_{1} e^{\frac{x}{2}}

\int 3^{2 - x} d x

\int_{\frac{8}{3}}^{4} 3 t - 8 d t

\frac{d}{d x} (ln (\frac{a x + 1}{a x - 1}))

d er i v a t i v e f (x) = 3 x + \frac{3}{x}

\frac{d}{d x} (\frac{e ^{x}}{1 - x})

What is the solution for 2y^'=e^{x/2}+y ?
The solution for 2y^'=e^{x/2}+y is y= x/(2e^{-x/2)}+c_{1}e^{x/2}