What is the solution for y^{'''}-2y^{''}+5y^'=0 ?

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

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y^{'''}-2y^{''}+5y^'=0

y^{^{'''}} - 2 y^{^{''}} + 5 y^{^{'}} = 0

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

Solution steps

y^{'''} (t) - 2 y^{''} (t) + 5 y^{'} (t) = 0

Solve linear ODE:

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

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

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

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

2 y^{^{''}} + 10 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 0

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

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