What is the solution for y^{'''}-5y^{''}+9y^'-5y=0 ?

The solution for y^{'''}-5y^{''}+9y^'-5y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(t)+c_{3}sin(t))

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y^{'''}-5y^{''}+9y^'-5y=0

y^{^{'''}} - 5 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

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

Solution steps

y^{'''} (t) - 5 y^{''} (t) + 9 y^{'} (t) - 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = - 4, y^{^{'}} (0) = - \frac{71}{5}

25 y^{^{''}} + 10 y^{^{'}} + 1 y = 0

3 y^{^{''}} - 13 y^{^{'}} + 14 y = 0

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

10 x^{^{''}} + 7 x^{^{'}} + x = 0

What is the solution for y^{'''}-5y^{''}+9y^'-5y=0 ?
The solution for y^{'''}-5y^{''}+9y^'-5y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(t)+c_{3}sin(t))