What is the solution for y^{''}-16y^'+73y=0 ?

The solution for y^{''}-16y^'+73y=0 is y=e^{8t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}-16y^'+73y=0

y^{^{''}} - 16 y^{^{'}} + 73 y = 0

y = e^{8 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) - 16 y^{'} (t) + 73 y = 0

Solve linear ODE:

y = e^{8 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{8 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y^{^{''}} - 8 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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

4 y^{^{''}} - 5 y^{^{'}} + y = 0

y^{^{''}} - (1 + 2) y^{^{'}} + 21 y = 0

y^{^{''}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 2 \cdot 2

What is the solution for y^{''}-16y^'+73y=0 ?
The solution for y^{''}-16y^'+73y=0 is y=e^{8t}(c_{1}cos(3t)+c_{2}sin(3t))