What is the solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(5t)+1/5 sin(5t))

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y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y = e^{- 2 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 2 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

y^{^{''}} + 6 y^{^{'}} + 7 y = 0, y (0) = 2, y^{^{'}} (0) = 0

- y^{^{''}} - 9 \cdot π^{2} y = 0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 324 x = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 23

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 0, y (0) = - 2, y^{^{'}} (0) = - 11

What is the solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(5t)+1/5 sin(5t))