What is the solution for y^{''}+12y^'+36y=0,y(-1)=2,y^'(-1)=5 ?

The solution for y^{''}+12y^'+36y=0,y(-1)=2,y^'(-1)=5 is y=19e^{-6t-6}+17e^{-6t-6}t

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y^{''}+12y^'+36y=0,y(-1)=2,y^'(-1)=5

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

y = 19 e^{- 6 t - 6} + 17 e^{- 6 t - 6} t

Solution steps

y^{''} (t) + 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 19 e^{- 6 t - 6} + 17 e^{- 6 t - 6} t

y = 19 e^{- 6 t - 6} + 17 e^{- 6 t - 6} t

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 8 y = 0

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

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y (0) = - \frac{9}{4}

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 7, y^{^{'}} (0) = 4

\frac{d ^{2} y}{d x ^{2}} = a \frac{d y}{d x}

What is the solution for y^{''}+12y^'+36y=0,y(-1)=2,y^'(-1)=5 ?
The solution for y^{''}+12y^'+36y=0,y(-1)=2,y^'(-1)=5 is y=19e^{-6t-6}+17e^{-6t-6}t