What is the solution for 4x^{''}+12x^'+9x=0 ?

The solution for 4x^{''}+12x^'+9x=0 is x=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t

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4x^{''}+12x^'+9x=0

4 x^{^{''}} + 12 x^{^{'}} + 9 x = 0

x = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

Solution steps

4 x^{''} (t) + 12 x^{'} (t) + 9 x = 0

Solve linear ODE:

x = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

x = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

y^{^{'''}} + 10 y^{^{''}} + 9 y^{^{'}} = 0

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

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y^{^{'}} (0) = - \frac{1}{3}

(D^{2} - 6 D + 25) y = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 2

What is the solution for 4x^{''}+12x^'+9x=0 ?
The solution for 4x^{''}+12x^'+9x=0 is x=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t