What is the solution for y^{''}-2y^'+20y=0 ?

The solution for y^{''}-2y^'+20y=0 is y=e^t(c_{1}cos(sqrt(19)t)+c_{2}sin(sqrt(19)t))

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y^{''}-2y^'+20y=0

y^{^{''}} - 2 y^{^{'}} + 20 y = 0

y = e^{t} (c_{1} cos (19 t) + c_{2} sin (19 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (19 t) + c_{2} sin (19 t))

y = e^{t} (c_{1} cos (19 t) + c_{2} sin (19 t))

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y^{^{''}} - 10 y^{^{'}} - 25 y = 0

55 y^{^{''''}} + 119 y^{^{''}} + 36 y = 0

y^{^{''}} - 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} + 51 y^{^{'}} + 36 y = 0, y (0) = - 5, y^{^{'}} (0) = - 20

What is the solution for y^{''}-2y^'+20y=0 ?
The solution for y^{''}-2y^'+20y=0 is y=e^t(c_{1}cos(sqrt(19)t)+c_{2}sin(sqrt(19)t))