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

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

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

linear

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

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 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^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

x^{^{''}} + \frac{1}{4} x^{^{'}} + x = 0

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 2

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 5

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