What is the solution for y^{''}+8y^'+20y=0,y(0)=1,y^'(0)=4 ?

The solution for y^{''}+8y^'+20y=0,y(0)=1,y^'(0)=4 is y=e^{-4t}(cos(2t)+4sin(2t))

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y^{''}+8y^'+20y=0,y(0)=1,y^'(0)=4

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y = e^{- 4 t} (cos (2 t) + 4 sin (2 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- 4 t} (cos (2 t) + 4 sin (2 t))

y = e^{- 4 t} (cos (2 t) + 4 sin (2 t))

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 7

x^{^{''}} - x^{^{'}} - 110 x = 0, x (0) = 14

y^{^{''}} - 6 y^{^{'}} - 40 y = 0

2 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + \frac{220}{55} y = 0

What is the solution for y^{''}+8y^'+20y=0,y(0)=1,y^'(0)=4 ?
The solution for y^{''}+8y^'+20y=0,y(0)=1,y^'(0)=4 is y=e^{-4t}(cos(2t)+4sin(2t))