What is the solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-13 ?

The solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-13 is y=e^{-4t}(4cos(5t)+3/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-13

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 4, y^{^{'}} (0) = - 13

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 20 y^{^{'}} + 136 y = 0

y^{^{'''}} + 13 y^{^{''}} + 45 y^{^{'}} + 33 y = 0

\frac{d ^{2} x}{d t ^{2}} - 49 x = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2 x}{3}}

y^{^{''''}} + 3 y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0

What is the solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-13 ?
The solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-13 is y=e^{-4t}(4cos(5t)+3/5 sin(5t))