What is the solution for y^{''}-6y^'+16y=0 ?

The solution for y^{''}-6y^'+16y=0 is y=e^{3t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))

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y^{''}-6y^'+16y=0

y^{^{''}} - 6 y^{^{'}} + 16 y = 0

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 5

3 y^{^{''}} + 2 y^{^{'}} + \frac{1}{4} y = 0

(D - 1) (D + 1) (D - 2) x = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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

What is the solution for y^{''}-6y^'+16y=0 ?
The solution for y^{''}-6y^'+16y=0 is y=e^{3t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))