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=49e^{-7t-28}+12e^{-7t-28}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 = 49 e^{- 7 t - 28} + 12 e^{- 7 t - 28} t

Solution steps

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

Solve linear ODE:

y = 49 e^{- 7 t - 28} + 12 e^{- 7 t - 28} t

y = 49 e^{- 7 t - 28} + 12 e^{- 7 t - 28} t

y^{^{'''}} - 49 y^{^{'}} = 0

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

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = e^{- \frac{π}{6}}

\frac{d ^{3} y}{d t ^{3}} - \frac{d ^{2} y}{d t ^{2}} = 0

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{18}{5}

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=49e^{-7t-28}+12e^{-7t-28}t