What is the solution for y^{''''}-y=e^{2t} ?

The solution for y^{''''}-y=e^{2t} is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)+1/15 e^{2t}

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y^{''''}-y=e^{2t}

y^{^{''''}} - y = e^{2 t}

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t) + \frac{1}{15} e^{2 t}

Solution steps

y^{(4)} (t) - y = e^{2 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t) + \frac{1}{15} e^{2 t}

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t) + \frac{1}{15} e^{2 t}

- \frac{1}{9} y^{^{''}} + \frac{2}{3} y^{^{'}} - y = \frac{t}{t + 1}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = t e^{- 3 t}

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

y^{^{''}} + y = 2 + t, y (0) = 1, y^{^{'}} (0) = 3

What is the solution for y^{''''}-y=e^{2t} ?
The solution for y^{''''}-y=e^{2t} is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)+1/15 e^{2t}