What is the solution for y^{''}-14y^'+65y=0 ?

The solution for y^{''}-14y^'+65y=0 is y=e^{7t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-14y^'+65y=0

y^{^{''}} - 14 y^{^{'}} + 65 y = 0

y = e^{7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 14 y^{'} (t) + 65 y = 0

Solve linear ODE:

y = e^{7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y^{^{''''}} - 81 y^{^{''}} = 0

3.5 y^{^{''}} - 5.6 y^{^{'}} + 2.2 y = 0

y^{^{'''}} - 2 y^{^{'}} + 4 y = 0

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 2 y = 0

\frac{d ^{2} y}{d t ^{2}} + 25 y = 0

What is the solution for y^{''}-14y^'+65y=0 ?
The solution for y^{''}-14y^'+65y=0 is y=e^{7t}(c_{1}cos(4t)+c_{2}sin(4t))