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

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{7}})

f (x) = 2 x^{3} - 4 x - 8 x + 6

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y^{^{'}} - a y + b = 0

\int_{6}^{8} \frac{56}{( x - 6 ) ^{3}} d 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