What is the solution for y^{''}-12y^'+40y=0 ?

The solution for y^{''}-12y^'+40y=0 is y=e^{6t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}-12y^'+40y=0

y^{^{''}} - 12 y^{^{'}} + 40 y = 0

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

Solution steps

y^{''} (t) - 12 y^{'} (t) + 40 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((\frac{x - 2}{x - π})^{- 3})

y = 2 x - 7

\frac{d}{d x} (a sin (w x))

x \to - \infty lim (\frac{11 x + 6}{4 - 6 x})

x = e^{2 y}, y = 0, x = e^{4}

What is the solution for y^{''}-12y^'+40y=0 ?
The solution for y^{''}-12y^'+40y=0 is y=e^{6t}(c_{1}cos(2t)+c_{2}sin(2t))