What is the solution for y^{''}+4y^'+y5=0 ?

The solution for y^{''}+4y^'+y5=0 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))

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y^{''}+4y^'+y5=0

y^{^{''}} + 4 y^{^{'}} + y 5 = 0

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t))

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = - 3, y^{^{'}} (0) = - 4

y^{^{''}} + 3 y = 0, y^{0} = 1, y^{^{'}} 0 = 3

\frac{1}{2} y^{^{''}} + 6 y^{^{'}} + 26 y = 0

y^{^{''}} - 9 y^{^{'}} = 0, y (0) = e^{3 x}

What is the solution for y^{''}+4y^'+y5=0 ?
The solution for y^{''}+4y^'+y5=0 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))