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

The solution for y^{''}+2y^'+5y=0,y(0)=-4,y^'(0)=3 is y=e^{-t}(-4cos(2t)-1/2 sin(2t))

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

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

y = e^{- t} (- 4 cos (2 t) - \frac{1}{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (- 4 cos (2 t) - \frac{1}{2} sin (2 t))

y = e^{- t} (- 4 cos (2 t) - \frac{1}{2} sin (2 t))

(D^{3} + 2 D^{2} - 15 D) y = 0

y^{^{''}} + b y^{^{'}} + 64 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + α y = 0

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

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

What is the solution for y^{''}+2y^'+5y=0,y(0)=-4,y^'(0)=3 ?
The solution for y^{''}+2y^'+5y=0,y(0)=-4,y^'(0)=3 is y=e^{-t}(-4cos(2t)-1/2 sin(2t))