What is the solution for y^{''''}+20y^{''}+64y=0 ?

The solution for y^{''''}+20y^{''}+64y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+c_{3}cos(4t)+c_{4}sin(4t)

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y^{''''}+20y^{''}+64y=0

y^{^{''''}} + 20 y^{^{''}} + 64 y = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (4 t) + c_{4} sin (4 t)

Solution steps

y^{(4)} (t) + 20 y^{''} (t) + 64 y = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (4 t) + c_{4} sin (4 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (4 t) + c_{4} sin (4 t)

\frac{d}{d x} (x^{2} arctan (6 x))

\int \frac{24 x ^{2}}{100 - x ^{2}} d x

\frac{d y}{d x} = 112 y x^{15}

\frac{d}{d x} (\frac{x ^{2}}{25})

\frac{d}{d x} (ln (\frac{x}{x ^{2} + 1}))

What is the solution for y^{''''}+20y^{''}+64y=0 ?
The solution for y^{''''}+20y^{''}+64y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+c_{3}cos(4t)+c_{4}sin(4t)