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

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

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

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

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

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

4 y^{^{''}} + 4 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 10, y^{^{'}} (0) = 10

y^{^{''}} - y^{^{'}} - 2 y = 0

(y^{^{''}} + y)^{2} = 0

y^{^{''}} + 6 y^{^{'}} + 18 y = 0, y (0) = 0, y^{^{'}} (0) = 0

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