What is the solution for y^{''}+12y^'+100y=0 ?

The solution for y^{''}+12y^'+100y=0 is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))

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y^{''}+12y^'+100y=0

y^{^{''}} + 12 y^{^{'}} + 100 y = 0

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

Solution steps

y^{''} (t) + 12 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

\int tan^{3} (2 t) sec^{4} (2 t) d t

\int \frac{4}{x ^{2} - 8 x + 12} d x

y^{^{''}} + y^{^{'}} - 6 y = 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

x + x^{- \frac{1}{3}}

\frac{\partial}{\partial x} (3 x^{2} y - 3 x + 3 y^{3})

What is the solution for y^{''}+12y^'+100y=0 ?
The solution for y^{''}+12y^'+100y=0 is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))