What is the solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-5 ?

The solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(3t)+1/3 sin(3t))

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y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-5

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

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

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

\int_{0}^{2 π} 5 d x

x \to 0 + lim (- x ln (x))

\int \frac{x - 2}{x ^{2} - 4 x + 3} d x

f (x) = ln (e^{x} + x e^{x})

x \to 0 lim (\frac{x ^{2} - 1}{x - 1})

What is the solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-5 ?
The solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(3t)+1/3 sin(3t))