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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{1}{( x ^{2} - 2 ) ( x ^{2} + x + 5 )}

y = 5 x cos (x), (π, - 5 π)

y = csc^{3} (2 x)

\frac{d}{d x} ((a x + b) e^{- x})

\int [x (x^{2} - 1) + csc^{2} (x)] d x

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