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

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

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y^{''}+2y^'+5y=0,y(0)=5,y^'(0)=-5

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

y = 5 e^{- t} cos (2 t)

Solution steps

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

Solve linear ODE:

y = 5 e^{- t} cos (2 t)

y = 5 e^{- t} cos (2 t)

x^{2} y^{^{''}} + 3 x y^{^{'}} + 4 y = 0

\frac{\partial}{\partial z} (sin (z^{3}) cos (x y))

f (x) = \frac{- 2 x}{x ^{5} - x - 1}

\int yz e^{x y} d x

y = x, (4, 2)

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