What is the solution for y^{''}-4y^'+68y=0 ?

The solution for y^{''}-4y^'+68y=0 is y=e^{2t}(c_{1}cos(8t)+c_{2}sin(8t))

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y^{''}-4y^'+68y=0

y^{^{''}} - 4 y^{^{'}} + 68 y = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 68 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 17 y = 0

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y (0) = 4, y^{^{'}} (0) = 17

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 5

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y^{^{'}} (0) = 8

What is the solution for y^{''}-4y^'+68y=0 ?
The solution for y^{''}-4y^'+68y=0 is y=e^{2t}(c_{1}cos(8t)+c_{2}sin(8t))