What is the solution for 3y^{''}+24y^'+48y=0,y(0)=8,y^'(0)=-33 ?

The solution for 3y^{''}+24y^'+48y=0,y(0)=8,y^'(0)=-33 is y=8e^{-4t}-e^{-4t}t

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3y^{''}+24y^'+48y=0,y(0)=8,y^'(0)=-33

3 y^{^{''}} + 24 y^{^{'}} + 48 y = 0, y (0) = 8, y^{^{'}} (0) = - 33

y = 8 e^{- 4 t} - e^{- 4 t} t

Solution steps

3 y^{''} (t) + 24 y^{'} (t) + 48 y = 0

Solve linear ODE:

y = 8 e^{- 4 t} - e^{- 4 t} t

y = 8 e^{- 4 t} - e^{- 4 t} t

y^{^{''}} - 4 y^{^{'}} + 29 y = 0, y (0) = 6, y^{^{'}} (0) = - 3

6 y^{^{''}} - 13 y^{^{'}} + 6 y = 0

y^{^{''}} + 5 y^{^{'}} + 8 y = 0

y^{^{'''}} - 19 y^{^{'}} + 30 y = 0

- \frac{d ^{2} y}{d x ^{2}} - \frac{y}{4} = 0

What is the solution for 3y^{''}+24y^'+48y=0,y(0)=8,y^'(0)=-33 ?
The solution for 3y^{''}+24y^'+48y=0,y(0)=8,y^'(0)=-33 is y=8e^{-4t}-e^{-4t}t