What is the solution for y^{''''}+2y^{''}+y=0 ?

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

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y^{''''}+2y^{''}+y=0

y^{^{''''}} + 2 y^{^{''}} + y = 0

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

Solution steps

y^{(4)} (t) + 2 y^{''} (t) + y = 0

Solve linear ODE:

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

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

e^{7 x}

f (x) = 3 x + 2

\frac{d}{d x} (20 (e^{- x} - e^{- 2 x}))

21 e^{t}

\int_{0}^{4} 2 π (y) (3 y + 4 - y^{2}) d y

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