What is the solution for y^{''}+14y^'+49y=0,y(-4)=-1,y(-4)=5 ?

The solution for y^{''}+14y^'+49y=0,y(-4)=-1,y(-4)=5 is y=e^{-7t-28}(-1+4e^{28}c_{2})+c_{2}e^{-7t}t

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y^{''}+14y^'+49y=0,y(-4)=-1,y(-4)=5

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (- 4) = - 1, y (- 4) = 5

y = e^{- 7 t - 28} (- 1 + 4 e^{28} c_{2}) + c_{2} e^{- 7 t} t

Solution steps

y^{''} (t) + 14 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = e^{- 7 t - 28} (- 1 + 4 e^{28} c_{2}) + c_{2} e^{- 7 t} t

y = e^{- 7 t - 28} (- 1 + 4 e^{28} c_{2}) + c_{2} e^{- 7 t} t

y^{^{'''}} - 5 y^{^{''}} + 4 y^{^{'}} - 20 y = 0

y^{^{''}} - 8 y^{^{'}} + 65 y = 0

y^{^{'''}} - 6 y^{^{''}} + 5 y^{^{'}} + 12 y = 0

y^{^{''}} + 32 y^{^{'}} + 224 y = 0

x^{^{''}} = x \cdot 5.3

What is the solution for y^{''}+14y^'+49y=0,y(-4)=-1,y(-4)=5 ?
The solution for y^{''}+14y^'+49y=0,y(-4)=-1,y(-4)=5 is y=e^{-7t-28}(-1+4e^{28}c_{2})+c_{2}e^{-7t}t