What is the solution for y^{''}+6y^'+13y=0,y(0)=-2,y^'(0)=0 ?

The solution for y^{''}+6y^'+13y=0,y(0)=-2,y^'(0)=0 is y=e^{-3t}(-2cos(2t)-3sin(2t))

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y^{''}+6y^'+13y=0,y(0)=-2,y^'(0)=0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = 0

y = e^{- 3 t} (- 2 cos (2 t) - 3 sin (2 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{- 3 t} (- 2 cos (2 t) - 3 sin (2 t))

y = e^{- 3 t} (- 2 cos (2 t) - 3 sin (2 t))

2 y^{^{''}} + 7 y^{^{'}} = 0

y^{^{''}} - 83 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

x^{^{''}} - 9 x = 0

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 4 x = 0, x (0) = 1

x^{^{''}} + a \cdot x = 0

What is the solution for y^{''}+6y^'+13y=0,y(0)=-2,y^'(0)=0 ?
The solution for y^{''}+6y^'+13y=0,y(0)=-2,y^'(0)=0 is y=e^{-3t}(-2cos(2t)-3sin(2t))