What is the solution for y^{''}+6y^'+5y=0,y(0)=3,y^'(0)=-1 ?

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

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y^{''}+6y^'+5y=0,y(0)=3,y^'(0)=-1

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y = (3 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 5 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = (3 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 5 t}

y = (3 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 5 t}

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 0

y^{^{'''}} - y^{^{''}} - 26 y^{^{'}} - 24 y = 0

y^{^{''}} - 2 y^{^{'}} + 20 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

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