What is the solution for x^{''}+4x^'+13x=te^t ?

The solution for x^{''}+4x^'+13x=te^t is x=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+(e^tt)/(18)-1/54 e^t

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x^{''}+4x^'+13x=te^t

x^{^{''}} + 4 x^{^{'}} + 13 x = t e^{t}

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{e ^{t} t}{18} - \frac{1}{54} e^{t}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 13 x = t e^{t}

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{e ^{t} t}{18} - \frac{1}{54} e^{t}

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{e ^{t} t}{18} - \frac{1}{54} e^{t}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 = 0

(\frac{d ^{2} y}{d x ^{2}} - 9 y) = e^{- 4 x} cos (e^{- x})

y^{^{'''}} + 8 y = 8 \cdot e^{- 2 \cdot t}

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

\frac{d ^{2} y}{d x ^{2}} + 36 y = 288 sin (6 x) + 96 cos (6 x)

What is the solution for x^{''}+4x^'+13x=te^t ?
The solution for x^{''}+4x^'+13x=te^t is x=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+(e^tt)/(18)-1/54 e^t