What is the solution for y^{''}-16y^'+68y=0 ?

The solution for y^{''}-16y^'+68y=0 is y=e^{8t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}-16y^'+68y=0

y^{^{''}} - 16 y^{^{'}} + 68 y = 0

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

Solution steps

y^{''} (t) - 16 y^{'} (t) + 68 y = 0

Solve linear ODE:

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

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

y^{^{''''}} + 4 y^{^{'''}} + 6 y^{^{''}} + 4 y^{^{'}} + y = 0

11 x^{^{''}} + 112.5 x = 0

\frac{y ^{^{''}}}{y} = a

\frac{d ^{2} y}{d x ^{2}} + 6 \cdot (\frac{d y}{d x}) + 8 \cdot y = 0

y^{^{''}} - 9 y = 0, y^{^{'}} (0) = 10

What is the solution for y^{''}-16y^'+68y=0 ?
The solution for y^{''}-16y^'+68y=0 is y=e^{8t}(c_{1}cos(2t)+c_{2}sin(2t))