What is the solution for y^{''}+y^'=3+e^x ?

The solution for y^{''}+y^'=3+e^x is y=c_{1}+c_{2}e^{-x}+3x+1/2 e^x

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y^{''}+y^'=3+e^x

y^{^{''}} + y^{^{'}} = 3 + e^{x}

y = c_{1} + c_{2} e^{- x} + 3 x + \frac{1}{2} e^{x}

Solution steps

y^{''} (x) + y^{'} (x) = 3 + e^{x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + 3 x + \frac{1}{2} e^{x}

y = c_{1} + c_{2} e^{- x} + 3 x + \frac{1}{2} e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2} e^{3 x} + 6

y^{^{''}} - y^{^{'}} - 2 y = 4 x^{2}, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 12

4 y^{^{''}} + 4 y^{^{'}} + y = e^{\frac{- x}{2}}

What is the solution for y^{''}+y^'=3+e^x ?
The solution for y^{''}+y^'=3+e^x is y=c_{1}+c_{2}e^{-x}+3x+1/2 e^x