What is the solution for y^{''}-30y^'+325y=0 ?

The solution for y^{''}-30y^'+325y=0 is y=e^{15t}(c_{1}cos(10t)+c_{2}sin(10t))

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y^{''}-30y^'+325y=0

y^{^{''}} - 30 y^{^{'}} + 325 y = 0

y = e^{15 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

Solution steps

y^{''} (t) - 30 y^{'} (t) + 325 y = 0

Solve linear ODE:

y = e^{15 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

y = e^{15 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

What is the solution for y^{''}-30y^'+325y=0 ?
The solution for y^{''}-30y^'+325y=0 is y=e^{15t}(c_{1}cos(10t)+c_{2}sin(10t))