What is the solution for y^{''}+6y^'+13y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+6y^'+13y=0,y(0)=1,y^'(0)=0 is y=e^{-3t}(cos(2t)+3/2 sin(2t))

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y^{''}+6y^'+13y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

9 y^{^{''}} - 16 y = 0

3 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = 0

y^{^{''}} + y^{^{'}} + 12 y = 0

9 y^{^{''}} + 36 y^{^{'}} + 100 y = 0, y (1) = 4, y^{^{'}} (1) = 9

y^{^{''}} = 8 y^{^{'}} - 15 y

What is the solution for y^{''}+6y^'+13y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+6y^'+13y=0,y(0)=1,y^'(0)=0 is y=e^{-3t}(cos(2t)+3/2 sin(2t))