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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{'''}} + y^{^{''}} - 6 y^{^{'}} + 4 y = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0

y^{^{''}} + 2 y + y = 0, y (0) = x e^{- x}

y^{^{''}} - 6 y^{^{'}} + 2 y = 0

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