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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 11

y^{^{''''}} + 24 y^{^{''}} = 0

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

y^{^{''}} + 5 y^{^{'}} - 3 y = 0

a^{2} \cdot \frac{d ^{2} y}{d x ^{2}} + a \cdot \frac{d y}{d x} - c \cdot y = 0

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