What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y(0)=-7 ?

The solution for y^{''}+4y^'+5y=0,y(0)=2,y(0)=-7 is y=e^{-2t}(2cos(t)+c_{2}sin(t))

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y^{''}+4y^'+5y=0,y(0)=2,y(0)=-7

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y (0) = - 7

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

Solution steps

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

Solve linear ODE:

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

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

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

x^{^{''}} + 6 x^{^{'}} + 3 x = 0

9 y^{^{''}} + 6 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{1}{3}

y^{^{''}} + 3 y^{^{'}} - 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

(D^{2} + 2 D + 1) y = 0

What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y(0)=-7 ?
The solution for y^{''}+4y^'+5y=0,y(0)=2,y(0)=-7 is y=e^{-2t}(2cos(t)+c_{2}sin(t))