What is the solution for y^{''}+12y^'+36y=0,y(-1)=5,y^'(-1)=3 ?

The solution for y^{''}+12y^'+36y=0,y(-1)=5,y^'(-1)=3 is y=38e^{-6t-6}+33e^{-6t-6}t

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y^{''}+12y^'+36y=0,y(-1)=5,y^'(-1)=3

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (- 1) = 5, y^{^{'}} (- 1) = 3

y = 38 e^{- 6 t - 6} + 33 e^{- 6 t - 6} t

Solution steps

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

Solve linear ODE:

y = 38 e^{- 6 t - 6} + 33 e^{- 6 t - 6} t

y = 38 e^{- 6 t - 6} + 33 e^{- 6 t - 6} t

4 y^{^{''}} - 19 y^{^{'}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 13.5

y^{^{''}} = 6 y^{^{'}} - 8 y, y^{^{'}} (0) = 2

(d^{3} - d^{2} + d - 1) y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 y = 0

y^{^{'''}} + 6 \cdot y^{^{''}} = 0

What is the solution for y^{''}+12y^'+36y=0,y(-1)=5,y^'(-1)=3 ?
The solution for y^{''}+12y^'+36y=0,y(-1)=5,y^'(-1)=3 is y=38e^{-6t-6}+33e^{-6t-6}t