What is the solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1 is y=e^{-3t}(3cos(2t)+5sin(2t))

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y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = e^{- 3 t} (3 cos (2 t) + 5 sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (3 cos (2 t) + 5 sin (2 t))

y = e^{- 3 t} (3 cos (2 t) + 5 sin (2 t))

y^{^{'''}} + 3 y^{^{''}} + 4 y^{^{'}} + 12 y = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

x^{^{''}} + \frac{1}{4} x^{^{'}} + x = 0

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}

What is the solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1 is y=e^{-3t}(3cos(2t)+5sin(2t))