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

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

y^{^{''''}} + 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))

\frac{d}{d x} ((\frac{10}{4 + x})^{\frac{1}{2}})

- \frac{1}{4} x^{4} + 6 x^{2}

\frac{d}{d x} (\frac{x ^{4}}{12})

f (x) = 4 x^{2} sin (x) + 2 x

cos (\frac{1 - e ^{6 x}}{1 + e ^{6 x}})

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