What is the solution for 2y^{''}+14y^'+12y=0,y(0)=1,y^'(0)=0 ?

The solution for 2y^{''}+14y^'+12y=0,y(0)=1,y^'(0)=0 is y=(1+1/5)e^{-t}-1/5 e^{-6t}

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2y^{''}+14y^'+12y=0,y(0)=1,y^'(0)=0

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{1}{5}) e^{- t} - \frac{1}{5} e^{- 6 t}

Solution steps

2 y^{''} (t) + 14 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = (1 + \frac{1}{5}) e^{- t} - \frac{1}{5} e^{- 6 t}

y = (1 + \frac{1}{5}) e^{- t} - \frac{1}{5} e^{- 6 t}

y^{^{''}} + 4 y^{^{'}} - 8 y = 0

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 20

\frac{d ^{2} x}{d t ^{2}} + 6 \cdot \frac{d x}{d t} + 3 \cdot x = 0, x (0) = 0

4 (\frac{d ^{2} y}{d x ^{2}}) + 4 (\frac{d y}{d x}) + y = 0

2500 y^{^{''}} + 81 y = 0

What is the solution for 2y^{''}+14y^'+12y=0,y(0)=1,y^'(0)=0 ?
The solution for 2y^{''}+14y^'+12y=0,y(0)=1,y^'(0)=0 is y=(1+1/5)e^{-t}-1/5 e^{-6t}