What is the solution for y^{''}+8y^'+41y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+8y^'+41y=0,y(0)=1,y^'(0)=2 is y=e^{-4t}(cos(5t)+6/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

\frac{3 y ^{^{''}} + 2 y ^{^{'}} + y}{4} = 0

16 y^{^{''}} + y = 0

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = 1.5

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{''}} + 3 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+8y^'+41y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+8y^'+41y=0,y(0)=1,y^'(0)=2 is y=e^{-4t}(cos(5t)+6/5 sin(5t))