What is the solution for 16y^{''}-8y^'+145y=0 ?

The solution for 16y^{''}-8y^'+145y=0 is y=e^{t/4}(c_{1}cos(3t)+c_{2}sin(3t))

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16y^{''}-8y^'+145y=0

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0

y = e^{\frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

16 y^{''} (t) - 8 y^{'} (t) + 145 y = 0

Solve linear ODE:

y = e^{\frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{\frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

What is the solution for 16y^{''}-8y^'+145y=0 ?
The solution for 16y^{''}-8y^'+145y=0 is y=e^{t/4}(c_{1}cos(3t)+c_{2}sin(3t))