What is the solution for y^{''}+8y^'+20y=0,y(0)=3,y^'(0)=-11 ?

The solution for y^{''}+8y^'+20y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(2t)+1/2 sin(2t))

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y^{''}+8y^'+20y=0,y(0)=3,y^'(0)=-11

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

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

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

\frac{d}{d x} (- 2 x^{3} + 12 x^{2} - 18 x)

\frac{\partial}{\partial x} (\frac{( 4 x + y )}{x - 3 y})

\frac{d}{d x} (arctan (\frac{x}{2}))

\int \frac{8 x ^{7} + 2 x}{x ^{8} + x ^{2}} d x

\frac{\partial}{\partial y} (\frac{2 x}{y ^{2}})

What is the solution for y^{''}+8y^'+20y=0,y(0)=3,y^'(0)=-11 ?
The solution for y^{''}+8y^'+20y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(2t)+1/2 sin(2t))