What is the solution for x^{'''}+x^{''}+3x^'-5x=0 ?

The solution for x^{'''}+x^{''}+3x^'-5x=0 is x=c_{1}e^t+e^{-t}(c_{2}cos(2t)+c_{3}sin(2t))

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x^{'''}+x^{''}+3x^'-5x=0

x^{^{'''}} + x^{^{''}} + 3 x^{^{'}} - 5 x = 0

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

Solution steps

x^{'''} (t) + x^{''} (t) + 3 x^{'} (t) - 5 x = 0

Solve linear ODE:

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

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

e^{x^{2} + 1}

f (x) = e^{\frac{4}{x}}

\frac{5 x}{x + 4}

\int (\frac{1}{1 + x}) d x

f (x) = e^{- a x} cos (b x - c)

What is the solution for x^{'''}+x^{''}+3x^'-5x=0 ?
The solution for x^{'''}+x^{''}+3x^'-5x=0 is x=c_{1}e^t+e^{-t}(c_{2}cos(2t)+c_{3}sin(2t))