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

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

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

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

y = e^{- 7 t - 28} (5 + 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} (5 + 4 e^{28} c_{2}) + c_{2} e^{- 7 t} t

y = e^{- 7 t - 28} (5 + 4 e^{28} c_{2}) + c_{2} e^{- 7 t} t

y^{^{''}} + y^{^{'}} + 9.25 y = 0, y (0) = 2, y^{^{'}} (0) = 8

y^{^{'''}} + 6 y^{^{''}} + 10 y^{^{'}} + 8 y = 0

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 0

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