What is the solution for y^{''}+8y^'+20y=0,y(0)=4,y^'(0)=-15 ?

The solution for y^{''}+8y^'+20y=0,y(0)=4,y^'(0)=-15 is y=e^{-4t}(4cos(2t)+1/2 sin(2t))

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y^{''}+8y^'+20y=0,y(0)=4,y^'(0)=-15

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

y = e^{- 4 t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

y = e^{- 4 t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

4 y^{^{''}} + π^{2} y = 0

3 y^{^{'''}} + 2 y^{^{''}} = 0

2 y^{^{''}} + 28 y^{^{'}} + 99 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 14 y^{^{'}} + 45 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 5, y^{^{'}} (0) = 28

What is the solution for y^{''}+8y^'+20y=0,y(0)=4,y^'(0)=-15 ?
The solution for y^{''}+8y^'+20y=0,y(0)=4,y^'(0)=-15 is y=e^{-4t}(4cos(2t)+1/2 sin(2t))