What is the solution for y^{'''}-5y^{''}+17y^'-13y=0 ?

The solution for y^{'''}-5y^{''}+17y^'-13y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}-5y^{''}+17y^'-13y=0

y^{^{'''}} - 5 y^{^{''}} + 17 y^{^{'}} - 13 y = 0

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

Solution steps

y^{'''} (t) - 5 y^{''} (t) + 17 y^{'} (t) - 13 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (1) = 1, y^{^{'}} (1) = 1

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = 0

y^{^{''}} + 22 y^{^{'}} + 122 y = 0, y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 0

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

What is the solution for y^{'''}-5y^{''}+17y^'-13y=0 ?
The solution for y^{'''}-5y^{''}+17y^'-13y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(3t)+c_{3}sin(3t))