What is the solution for y^{''}+14y^'+48y=0,y(0)=0,y^'(0)=-6 ?

The solution for y^{''}+14y^'+48y=0,y(0)=0,y^'(0)=-6 is y=-3e^{-6t}+3e^{-8t}

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y^{''}+14y^'+48y=0,y(0)=0,y^'(0)=-6

y^{^{''}} + 14 y^{^{'}} + 48 y = 0, y (0) = 0, y^{^{'}} (0) = - 6

y = - 3 e^{- 6 t} + 3 e^{- 8 t}

Solution steps

y^{''} (t) + 14 y^{'} (t) + 48 y = 0

Solve linear ODE:

y = - 3 e^{- 6 t} + 3 e^{- 8 t}

y = - 3 e^{- 6 t} + 3 e^{- 8 t}

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 5

2 y^{^{''}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = 35

y^{^{''}} + 9 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = - 45

y^{^{''}} + \frac{33}{6} y^{^{'}} + 128 y = 0

15 y^{^{''}} - 16 y^{^{'}} - 7 y = 0

What is the solution for y^{''}+14y^'+48y=0,y(0)=0,y^'(0)=-6 ?
The solution for y^{''}+14y^'+48y=0,y(0)=0,y^'(0)=-6 is y=-3e^{-6t}+3e^{-8t}