What is the solution for (D^3+D-10)y=29e^{4x} ?

The solution for (D^3+D-10)y=29e^{4x} is y=c_{1}e^{2x}+e^{-x}(c_{2}cos(2x)+c_{3}sin(2x))+1/2 e^{4x}

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(D^3+D-10)y=29e^{4x}

(D^{3} + D - 10) y = 29 e^{4 x}

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

Solution steps

(D^{3} + D - 10) y = 29 e^{4 x}

Solve linear ODE:

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

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

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

y^{^{''''}} - 4 y^{^{''}} = x^{2}

y^{^{''}} + 2 y^{^{'}} + 5 y = 12 e^{- t} cos (2 t)

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3}

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

What is the solution for (D^3+D-10)y=29e^{4x} ?
The solution for (D^3+D-10)y=29e^{4x} is y=c_{1}e^{2x}+e^{-x}(c_{2}cos(2x)+c_{3}sin(2x))+1/2 e^{4x}