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

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

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

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

y = \frac{1}{3} + 3 - \frac{e ^{12}}{3} e^{- 12 t} - 3 t

Solution steps

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

Solve linear ODE:

y = \frac{1}{3} + 3 - \frac{e ^{12}}{3} e^{- 12 t} - 3 t

y = \frac{1}{3} + 3 - \frac{e ^{12}}{3} e^{- 12 t} - 3 t

y^{^{'}} + y = (x + 1)^{2}

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