What is the solution for 9(d^2y)/(dt^2)-24(dy)/(dt)+16y=0 ?

The solution for 9(d^2y)/(dt^2)-24(dy)/(dt)+16y=0 is y=c_{1}e^{(4t)/3}+c_{2}e^{(4t)/3}t

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9(d^2y)/(dt^2)-24(dy)/(dt)+16y=0

9 \frac{d ^{2} y}{d t ^{2}} - 24 \frac{d y}{d t} + 16 y = 0

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

Solution steps

9 \frac{d ^{2} y}{d t ^{2}} - 24 \frac{d y}{d 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^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

(4 d^{4} - 8 d^{3} - 4 d^{2} + 8 d) y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

4 y^{^{''}} + π^{2} y = 0

What is the solution for 9(d^2y)/(dt^2)-24(dy)/(dt)+16y=0 ?
The solution for 9(d^2y)/(dt^2)-24(dy)/(dt)+16y=0 is y=c_{1}e^{(4t)/3}+c_{2}e^{(4t)/3}t