What is the solution for y^{'''}-7y^{''}+19y^'-13y=0 ?

The solution for y^{'''}-7y^{''}+19y^'-13y=0 is y=c_{1}e^t+e^{3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}-7y^{''}+19y^'-13y=0

y^{^{'''}} - 7 y^{^{''}} + 19 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) - 7 y^{''} (t) + 19 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))

y^{^{'''}} - 3 y^{^{''}} - 9 y^{^{'}} + 27 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} (0) = x e^{- x}

8 \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 9 y = 0

y^{^{'''}} + 4 y^{^{''}} + 3 y^{^{'}} = 0

y^{^{''}} = a y

What is the solution for y^{'''}-7y^{''}+19y^'-13y=0 ?
The solution for y^{'''}-7y^{''}+19y^'-13y=0 is y=c_{1}e^t+e^{3t}(c_{2}cos(2t)+c_{3}sin(2t))