What is the solution for y^{''}-2y^'+24y=0 ?

The solution for y^{''}-2y^'+24y=0 is y=e^t(c_{1}cos(sqrt(23)t)+c_{2}sin(sqrt(23)t))

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y^{''}-2y^'+24y=0

y^{^{''}} - 2 y^{^{'}} + 24 y = 0

y = e^{t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 24 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{t} (c_{1} cos (23 t) + c_{2} sin (23 t))

35 y^{^{''''}} + 153 y^{^{''}} + 100 y = 0

y^{^{''}} - 8 y^{^{'}} - 9 y = 0, y (0) = 0, y^{^{'}} (0) = 4

(D^{4} + 2 D^{3} + D^{2}) y = 0

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

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

What is the solution for y^{''}-2y^'+24y=0 ?
The solution for y^{''}-2y^'+24y=0 is y=e^t(c_{1}cos(sqrt(23)t)+c_{2}sin(sqrt(23)t))