What is the solution for y^{''}+8y^'+20y=0,y(0)=2,y^'(0)=-7 ?

The solution for y^{''}+8y^'+20y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(2t)+1/2 sin(2t))

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y^{''}+8y^'+20y=0,y(0)=2,y^'(0)=-7

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

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

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

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

y^{^{''''}} + 8 y^{^{'''}} + 16 y^{^{''}} = 0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0

What is the solution for y^{''}+8y^'+20y=0,y(0)=2,y^'(0)=-7 ?
The solution for y^{''}+8y^'+20y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(2t)+1/2 sin(2t))