What is the solution for y^{''}+2y=1+e^x ?

The solution for y^{''}+2y=1+e^x is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)+1/2+1/3 e^x

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y^{''}+2y=1+e^x

y^{^{''}} + 2 y = 1 + e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2} + \frac{1}{3} e^{x}

Solution steps

y^{''} (x) + 2 y = 1 + e^{x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2} + \frac{1}{3} e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2} + \frac{1}{3} e^{x}

\frac{d ^{4} y}{d t ^{4}} - 4 y^{^{''}} = t^{2} + e^{t}

y^{^{''}} - 6 y^{^{'}} + 25 y = sin (\frac{t}{2}) + e^{- t} t

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} = 3 e^{x} + 2 x e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = t, y (0) = 1, y^{^{'}} (0) = 1

(D^{2} + 4 D + 5) y = 50 x + 13 e^{x}

What is the solution for y^{''}+2y=1+e^x ?
The solution for y^{''}+2y=1+e^x is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)+1/2+1/3 e^x