What is the solution for y^{''}-12y^'+136y=0 ?

The solution for y^{''}-12y^'+136y=0 is y=e^{6t}(c_{1}cos(10t)+c_{2}sin(10t))

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y^{''}-12y^'+136y=0

y^{^{''}} - 12 y^{^{'}} + 136 y = 0

y = e^{6 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

Solution steps

y^{''} (t) - 12 y^{'} (t) + 136 y = 0

Solve linear ODE:

y = e^{6 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

y = e^{6 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

256 y^{^{''''}} = 16 y

y^{^{''}} - 7 y^{^{'}} - 4 y = 0

y^{^{''}} - 15 y^{^{'}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = - 55

y^{^{''}} + 10 y^{^{'}} + \frac{8}{3} y = 0, y (0) = 0, y^{^{'}} (0) = - 5

15 y^{^{''''}} + 32 y^{^{''}} + 9 y = 0

What is the solution for y^{''}-12y^'+136y=0 ?
The solution for y^{''}-12y^'+136y=0 is y=e^{6t}(c_{1}cos(10t)+c_{2}sin(10t))