What is the solution for 4y^{''''}+8y^{''}+4y=0 ?

The solution for 4y^{''''}+8y^{''}+4y=0 is y=c_{1}cos(t)+c_{2}sin(t)+t(c_{3}cos(t)+c_{4}sin(t))

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4y^{''''}+8y^{''}+4y=0

4 y^{^{''''}} + 8 y^{^{''}} + 4 y = 0

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

Solution steps

4 y^{(4)} (t) + 8 y^{''} (t) + 4 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{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 21 y = 0

(D^{2} + 2)^{3} (D^{2} - 3)^{2} y = 0

y^{^{'''}} + 3 y^{^{''}} - 5 y^{^{'}} - 39 y = 0, y (0) = e^{3 x}

(D^{2} + 1)^{3} y = 0

y^{^{''}} - y^{^{'}} - 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for 4y^{''''}+8y^{''}+4y=0 ?
The solution for 4y^{''''}+8y^{''}+4y=0 is y=c_{1}cos(t)+c_{2}sin(t)+t(c_{3}cos(t)+c_{4}sin(t))