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

The solution for 16y^{''}+24y^'+9y=0 is y=c_{1}e^{-(3t)/4}+c_{2}e^{-(3t)/4}t

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

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

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- \frac{3 t}{4}} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- \frac{3 t}{4}} t

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- \frac{3 t}{4}} t

\int 6 r 4 - r^{2} d r

\frac{d y}{d x} = e^{x + 2 y}

\frac{d y}{d x} = y (y + 2)

x \to 11 lim (\frac{2 x ^{2} - 12 x - 110}{x ^{2} - 121})

f (x) = 2 \cdot x

What is the solution for 16y^{''}+24y^'+9y=0 ?
The solution for 16y^{''}+24y^'+9y=0 is y=c_{1}e^{-(3t)/4}+c_{2}e^{-(3t)/4}t