What is the solution for y^{''}-16y^'+100y=0 ?

The solution for y^{''}-16y^'+100y=0 is y=e^{8t}(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}-16y^'+100y=0

y^{^{''}} - 16 y^{^{'}} + 100 y = 0

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

Solution steps

y^{''} (t) - 16 y^{'} (t) + 100 y = 0

Solve linear ODE:

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

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

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f (x) = \frac{e ^{3 x}}{x ^{15}}

x \to + 2 - lim (x - 2)

\int \frac{e ^{u}}{( 8 - e ^{u} ) ^{2}} d u

(1 + 5 x^{2}) (x - x^{2})

What is the solution for y^{''}-16y^'+100y=0 ?
The solution for y^{''}-16y^'+100y=0 is y=e^{8t}(c_{1}cos(6t)+c_{2}sin(6t))