What is the solution for 8y^{''}-8y^'+100y=0 ?

The solution for 8y^{''}-8y^'+100y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))

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8y^{''}-8y^'+100y=0

8 y^{^{''}} - 8 y^{^{'}} + 100 y = 0

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

Solution steps

8 y^{''} (t) - 8 y^{'} (t) + 100 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 9 y^{^{''}} + 18 y = 0, y (0) = 9

y^{^{''}} + a y^{^{'}} + b y = 0, y (0) = 2 cos (3 x)

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

y^{^{''}} + 2 y^{^{'}} + a y = 0

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

What is the solution for 8y^{''}-8y^'+100y=0 ?
The solution for 8y^{''}-8y^'+100y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))