What is the solution for y^{''}-4y^'+9y=0 ?

The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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y^{''}-4y^'+9y=0

y^{^{''}} - 4 y^{^{'}} + 9 y = 0

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y^{^{''}} + 22 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 25

y^{^{''}} + \frac{2}{3} y^{^{'}} + \frac{1}{9} y = 0

(D^{2} - 2 D + 5) y = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0

y^{^{''''}} + 2 y^{^{'''}} + 4 y^{^{''}} - 2 y^{^{'}} - 5 y = 0

What is the solution for y^{''}-4y^'+9y=0 ?
The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))