What is the solution for 2y^{''''}+4y^{''}+2y=0 ?

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

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2y^{''''}+4y^{''}+2y=0

2 y^{^{''''}} + 4 y^{^{''}} + 2 y = 0

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

Solution steps

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

121 y^{^{''}} + 64 y^{^{'}} = 0

y^{^{''}} + 8 y^{^{'}} + 52 y = 0, y (0) = 6, y^{^{'}} (0) = 7

\frac{d ^{2} x}{d t ^{2}} = - π^{2} x

u^{^{''}} + 0.25 u^{^{'}} + H (t) = 0

y^{^{'''}} - 23 y^{^{''}} + 160 y^{^{'}} - 300 y = 0

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