What is the solution for y^{''}+2y^'+5y=0 ?

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

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y^{''}+2y^'+5y=0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\int \frac{1}{( x + 4 ) ^{3}} d x

y^{^{''}} - 6 y^{^{'}} + 7 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\int \frac{1}{x ( 3 x + 1 )} d x

\frac{d}{d t} (sin^{4} (t))

(t^{2} - 20 t + 75) \frac{d y}{d t} = y

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