What is the solution for y^{''}-18y^'+97y=0 ?

The solution for y^{''}-18y^'+97y=0 is y=e^{9t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-18y^'+97y=0

y^{^{''}} - 18 y^{^{'}} + 97 y = 0

y = e^{9 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 18 y^{'} (t) + 97 y = 0

Solve linear ODE:

y = e^{9 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{9 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\frac{d ^{3} y}{d x ^{3}} + 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{''}} + \frac{16}{3} y = 0

4 y^{^{''}} + 22 y^{^{'}} + 18 y = 0, y (0) = 3, y^{^{'}} (0) = 0

2 y^{^{''}} - 5 y - 3 y = 0

y^{^{''}} + 10.5 y = 0

What is the solution for y^{''}-18y^'+97y=0 ?
The solution for y^{''}-18y^'+97y=0 is y=e^{9t}(c_{1}cos(4t)+c_{2}sin(4t))