What is the solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(4t)+1/4 sin(4t))

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y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

y^{^{'''}} - 4 y^{^{''}} - 11 y^{^{'}} - 6 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 8, y (1) = 6

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0, y (0) = 0

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-3 is y=e^{-2t}(2cos(4t)+1/4 sin(4t))