What is the solution for y^{''''}-(-5)y^{''}+4y=0 ?

The solution for y^{''''}-(-5)y^{''}+4y=0 is y=c_{1}cos(t)+c_{2}sin(t)+c_{3}cos(2t)+c_{4}sin(2t)

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y^{''''}-(-5)y^{''}+4y=0

y^{^{''''}} - (- 5) y^{^{''}} + 4 y = 0

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

Solution steps

y^{(4)} (t) - (- 5) y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

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

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 1, y^{^{'}} (0) = 10

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

4 y^{^{''}} + 324 y = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

What is the solution for y^{''''}-(-5)y^{''}+4y=0 ?
The solution for y^{''''}-(-5)y^{''}+4y=0 is y=c_{1}cos(t)+c_{2}sin(t)+c_{3}cos(2t)+c_{4}sin(2t)