What is the solution for 16y^{''}+24y^'+9=0 ?

The solution for 16y^{''}+24y^'+9=0 is y=c_{1}+c_{2}e^{-(3t)/2}-(3t)/8

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16y^{''}+24y^'+9=0

16 y^{^{''}} + 24 y^{^{'}} + 9 = 0

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3 t}{8}

Solution steps

16 y^{''} (t) + 24 y^{'} (t) + 9 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3 t}{8}

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3 t}{8}

(D^{3} + D^{2} - 4 D - 4) y = 3 e^{- x} - 4 x - 6

y^{^{''}} + 2 y^{^{'}} + 5 y = 30 x^{3} + 2

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

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

What is the solution for 16y^{''}+24y^'+9=0 ?
The solution for 16y^{''}+24y^'+9=0 is y=c_{1}+c_{2}e^{-(3t)/2}-(3t)/8