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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 4 t, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 4 y = 9 t

y^{^{''}} + 4 y = 3 csc (x)

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 2 x e^{- x}

y^{^{''}} + 4 y = cos^{2} (2 x)

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