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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 1

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

\frac{d ^{2} y}{d x ^{2}} = (4 y)

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = 1

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