What is the solution for y^{''}+4y^'+13y=0,y(0)=1,y^'(0)=2 ?

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

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y^{''}+4y^'+13y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- 2 t} (cos (3 t) + \frac{4}{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{4}{3} sin (3 t))

y = e^{- 2 t} (cos (3 t) + \frac{4}{3} sin (3 t))

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 15 y = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

20 y^{^{''}} + 9 y^{^{'}} + y = 0

x^{^{''}} + 4 x = 0, x (0) = 8, x^{^{'}} (0) = - 13

\frac{d ^{2} y}{d x ^{2}} = n y

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