What is the solution for x^{'''}+4x^{''}+13x^'=0 ?

The solution for x^{'''}+4x^{''}+13x^'=0 is x=c_{1}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))

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x^{'''}+4x^{''}+13x^'=0

x^{^{'''}} + 4 x^{^{''}} + 13 x^{^{'}} = 0

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

Solution steps

x^{'''} (t) + 4 x^{''} (t) + 13 x^{'} (t) = 0

Solve linear ODE:

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

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

(d^{4} - 5 d^{3} + 5 d^{2} + 5 d - 6) y = 0

14 y^{^{''''}} + 95 y^{^{''}} + 144 y = 0

y^{^{''}} = 5 y

5 \frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0, x (0) = 2

\frac{d ^{2} f}{d x ^{2}} = 2 f (x)

What is the solution for x^{'''}+4x^{''}+13x^'=0 ?
The solution for x^{'''}+4x^{''}+13x^'=0 is x=c_{1}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))