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

The solution for y^{''''}+2*y^{''}+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^{''''}+2*y^{''}+y=0

y^{^{''''}} + 2 \cdot 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 \cdot 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))

\frac{\partial}{\partial y} (2 x y + 2 x)

\frac{d}{d x} (sin^{2} (8 x))

\frac{d y}{d x} + x y = x y^{2}

\frac{d}{d x} (- cos (cos (x)) sin (x))

\int x^{3} sin^{2} (x) d x

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