What is the solution for 1/2 y^{''}+3y^'+4y=0,y(0)=0,y^'(0)=-5/3 ?

The solution for 1/2 y^{''}+3y^'+4y=0,y(0)=0,y^'(0)=-5/3 is y=-5/6 e^{-2t}+5/6 e^{-4t}

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1/2 y^{''}+3y^'+4y=0,y(0)=0,y^'(0)=-5/3

\frac{1}{2} y^{^{''}} + 3 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - \frac{5}{3}

y = - \frac{5}{6} e^{- 2 t} + \frac{5}{6} e^{- 4 t}

Solution steps

\frac{1}{2} y^{''} (t) + 3 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = - \frac{5}{6} e^{- 2 t} + \frac{5}{6} e^{- 4 t}

y = - \frac{5}{6} e^{- 2 t} + \frac{5}{6} e^{- 4 t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 4, y^{^{'}} (0) = 14

y^{^{''}} + y = 0, y (0) = 11, y^{^{'}} (0) = 33

\frac{1}{20} y^{^{''}} + 3 y^{^{'}} + 80 y = 0

y^{^{''}} - 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 3

7 x^{^{''}} + 3 x^{^{''}} + 7 x = 0, x (0) = - 4, x^{^{'}} (0) = 0

What is the solution for 1/2 y^{''}+3y^'+4y=0,y(0)=0,y^'(0)=-5/3 ?
The solution for 1/2 y^{''}+3y^'+4y=0,y(0)=0,y^'(0)=-5/3 is y=-5/6 e^{-2t}+5/6 e^{-4t}