What is the solution for y^{''}-8y^'+52y=0 ?

The solution for y^{''}-8y^'+52y=0 is y=e^{4t}(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}-8y^'+52y=0

y^{^{''}} - 8 y^{^{'}} + 52 y = 0

y = e^{4 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 52 y = 0

Solve linear ODE:

y = e^{4 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{4 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

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What is the solution for y^{''}-8y^'+52y=0 ?
The solution for y^{''}-8y^'+52y=0 is y=e^{4t}(c_{1}cos(6t)+c_{2}sin(6t))