What is the solution for y^{''}+4y^'+20y=0,y(0)=3,y^'(0)=-5 ?

The solution for y^{''}+4y^'+20y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(4t)+1/4 sin(4t))

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y^{''}+4y^'+20y=0,y(0)=3,y^'(0)=-5

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

y = e^{- 2 t} (3 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} (3 cos (4 t) + \frac{1}{4} sin (4 t))

y = e^{- 2 t} (3 cos (4 t) + \frac{1}{4} sin (4 t))

4 y^{^{''}} - 2 y^{^{'}} = 0

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

x^{^{''}} - 2 x^{^{'}} + 2 x = 0, x (0) = 1, x^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+4y^'+20y=0,y(0)=3,y^'(0)=-5 ?
The solution for y^{''}+4y^'+20y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(4t)+1/4 sin(4t))