What is the solution for y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=1 ?

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

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y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} ((x + 1)^{s i n (x)})

2 \frac{d y}{d x} = y - e^{4 x}

f (x) = 6 x (6.6)

\frac{\partial}{\partial y} (x sin (x + 2 y))

y^{^{''}} - 2 y^{^{'}} - 8 y = 7 - e^{4 x}

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