What is the solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=0 ?

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

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y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{( 1 + y ^{2} ) arctan ( y )} d y

y = \frac{1}{14 x ^{2}}, y = x, y = \frac{x}{12}

n = 1 \sum \infty n^{2} + 1

2 x y - y^{2} = - 1

\frac{d}{d x} (cos^{4} (4 x^{2} + 1))

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