What is the solution for y^{''}+12y^'+36y=0,y(0)=-2,y^'(0)=16 ?

The solution for y^{''}+12y^'+36y=0,y(0)=-2,y^'(0)=16 is y=-2e^{-6t}+4e^{-6t}t

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y^{''}+12y^'+36y=0,y(0)=-2,y^'(0)=16

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

y = - 2 e^{- 6 t} + 4 e^{- 6 t} t

Solution steps

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

Solve linear ODE:

y = - 2 e^{- 6 t} + 4 e^{- 6 t} t

y = - 2 e^{- 6 t} + 4 e^{- 6 t} t

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

6 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

x^{^{''}} + 2 x^{^{'}} + 9 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} = 0

What is the solution for y^{''}+12y^'+36y=0,y(0)=-2,y^'(0)=16 ?
The solution for y^{''}+12y^'+36y=0,y(0)=-2,y^'(0)=16 is y=-2e^{-6t}+4e^{-6t}t