What is the solution for (d^4y)/(dt^4)+2y^{''}+y=0 ?

The solution for (d^4y)/(dt^4)+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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(d^4y)/(dt^4)+2y^{''}+y=0

\frac{d ^{4} y}{d t ^{4}} + 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))

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{''}} + y^{^{'}} = 0, y (0) = 2

0.05 y^{^{''}} + 2 y^{^{'}} + \frac{1}{0.01} y = 0

3 y^{^{''}} - 2 y^{^{'}} - y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + y = 0

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