What is the solution for x^{''}+6x^'+9x=0,x(0)=a,x^'(0)=b ?

The solution for x^{''}+6x^'+9x=0,x(0)=a,x^'(0)=b is x=ae^{-3t}+(b+3a)te^{-3t}

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x^{''}+6x^'+9x=0,x(0)=a,x^'(0)=b

x^{^{''}} + 6 x^{^{'}} + 9 x = 0, x (0) = a, x^{^{'}} (0) = b

x = a e^{- 3 t} + (b + 3 a) t e^{- 3 t}

Solution steps

x^{''} (t) + 6 x^{'} (t) + 9 x = 0

Solve linear ODE:

x = a e^{- 3 t} + (b + 3 a) t e^{- 3 t}

x = a e^{- 3 t} + (b + 3 a) t e^{- 3 t}

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

9 (\frac{d ^{2} y}{d x ^{2}}) + 12 (\frac{d y}{d x}) + 4 y = 0

2 y^{^{''}} + 4 y^{^{'}} + k y = 0

\frac{d ^{2} y}{d x ^{2}} (4) + \frac{d y}{d x} (5) + y = 0

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = - 3

What is the solution for x^{''}+6x^'+9x=0,x(0)=a,x^'(0)=b ?
The solution for x^{''}+6x^'+9x=0,x(0)=a,x^'(0)=b is x=ae^{-3t}+(b+3a)te^{-3t}