What is the solution for y^{'''}+5y^{''}+7y^'-13y=0 ?

The solution for y^{'''}+5y^{''}+7y^'-13y=0 is y=c_{1}e^t+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+5y^{''}+7y^'-13y=0

y^{^{'''}} + 5 y^{^{''}} + 7 y^{^{'}} - 13 y = 0

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

Solution steps

y^{'''} (t) + 5 y^{''} (t) + 7 y^{'} (t) - 13 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((x^{2} + 2 x - 1) - (2 x^{2} - 3 x + 6))

\int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{d y}{d x} = (x + y - 2)^{2}

\frac{d ^{2}}{d x ^{2}} (3 x - 7)

2 x^{2} + 5 x + 6, (- 2, 0)

What is the solution for y^{'''}+5y^{''}+7y^'-13y=0 ?
The solution for y^{'''}+5y^{''}+7y^'-13y=0 is y=c_{1}e^t+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))