What is the solution for y^{''}+4y^'+13y=0,y(0)=-1,y^'(0)=1 ?

The solution for y^{''}+4y^'+13y=0,y(0)=-1,y^'(0)=1 is y=e^{-2t}(-cos(3t)-1/3 sin(3t))

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y^{''}+4y^'+13y=0,y(0)=-1,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 1

y = e^{- 2 t} (- cos (3 t) - \frac{1}{3} sin (3 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{- 2 t} (- cos (3 t) - \frac{1}{3} sin (3 t))

y = e^{- 2 t} (- cos (3 t) - \frac{1}{3} sin (3 t))

\int \frac{1}{( x - 6 ) ( x - 2 )} d x

f (x) = 4 sec (x), at \frac{π}{3}

\int (3 x^{2} (2 x^{3} - 1)) d x

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 4, y^{^{'}} (0) = - 1

e^{x} - 1

What is the solution for y^{''}+4y^'+13y=0,y(0)=-1,y^'(0)=1 ?
The solution for y^{''}+4y^'+13y=0,y(0)=-1,y^'(0)=1 is y=e^{-2t}(-cos(3t)-1/3 sin(3t))