What is the solution for y^{''}-6y^'+32y=0 ?

The solution for y^{''}-6y^'+32y=0 is y=e^{3t}(c_{1}cos(sqrt(23)t)+c_{2}sin(sqrt(23)t))

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y^{''}-6y^'+32y=0

y^{^{''}} - 6 y^{^{'}} + 32 y = 0

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = sin (3 x)

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

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

y^{^{''}} - 12 y^{^{'}} + 45 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

10 y^{^{''}} + 45 y^{^{'}} + 2500 y = 0

What is the solution for y^{''}-6y^'+32y=0 ?
The solution for y^{''}-6y^'+32y=0 is y=e^{3t}(c_{1}cos(sqrt(23)t)+c_{2}sin(sqrt(23)t))