What is the solution for y^{''}+5y^'+4y=0,y(0)=6,y^'(0)=2 ?

The solution for y^{''}+5y^'+4y=0,y(0)=6,y^'(0)=2 is y=(6+8/3)e^{-t}-8/3 e^{-4t}

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y^{''}+5y^'+4y=0,y(0)=6,y^'(0)=2

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = 2

y = (6 + \frac{8}{3}) e^{- t} - \frac{8}{3} e^{- 4 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (6 + \frac{8}{3}) e^{- t} - \frac{8}{3} e^{- 4 t}

y = (6 + \frac{8}{3}) e^{- t} - \frac{8}{3} e^{- 4 t}

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''}} + n^{2} y = 0

What is the solution for y^{''}+5y^'+4y=0,y(0)=6,y^'(0)=2 ?
The solution for y^{''}+5y^'+4y=0,y(0)=6,y^'(0)=2 is y=(6+8/3)e^{-t}-8/3 e^{-4t}