What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(t)+sin(t))

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y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} - 8 y^{^{'}} + 6 y = 0

y^{^{'''}} + 4 y^{^{''}} - 11 y^{^{'}} + 6 y = 0

3 y^{^{''}} - 4 y^{^{'}} + 7 y = 0

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = e^{t}

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(t)+sin(t))