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

The solution for y^{''}+12y^'+36=0 is y=c_{1}+c_{2}e^{-12t}-3t

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y^{''}+12y^'+36=0

y^{^{''}} + 12 y^{^{'}} + 36 = 0

y = c_{1} + c_{2} e^{- 12 t} - 3 t

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 12 t} - 3 t

y = c_{1} + c_{2} e^{- 12 t} - 3 t

\frac{\partial}{\partial x} (y e^{x} + x ln (z))

t y^{^{'}} = 4 t^{2} - 2 y

y^{^{''}} x = y^{^{'}}

\frac{\partial}{\partial u} (u v^{2} + w^{3})

\frac{\partial}{\partial x} (x^{2} y e^{x y})

What is the solution for y^{''}+12y^'+36=0 ?
The solution for y^{''}+12y^'+36=0 is y=c_{1}+c_{2}e^{-12t}-3t