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))

\int - 2 e^{- 2 x} d x

y^{^{''}} + 16 y = - 8 sin (4 t)

\frac{4}{x ^{5}}

\int csc (π) x cot (π) x d x

7 x - 6 x (1.1)

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))