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))

y^{^{'}} = 5 y + e^{- 2 x} y^{- 2}, y (0) = 2

f (x) = x^{10} e^{x}

3^{(x^{5})}

n = 0 \sum \infty 1 + n

y^{^{'}} + 2 y = 4, y (0) = 4

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))