What is the solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-11 ?

The solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(5t)+1/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-11

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

y = e^{- 4 t} (3 cos (5 t) + \frac{1}{5} sin (5 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

y = e^{- 4 t} (3 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 4 t} (3 cos (5 t) + \frac{1}{5} sin (5 t))

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

2 x^{^{''}} + 2 x^{^{'}} + 7 x = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 20, y^{^{'}} (0) = 0

x^{^{''}} + 6 x^{^{'}} + 13 x = 0

What is the solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-11 ?
The solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(5t)+1/5 sin(5t))