What is the solution for 324y^{''}-396y^'+121y=0 ?

The solution for 324y^{''}-396y^'+121y=0 is y=c_{1}e^{(11t)/(18)}+c_{2}e^{(11t)/(18)}t

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324y^{''}-396y^'+121y=0

324 y^{^{''}} - 396 y^{^{'}} + 121 y = 0

y = c_{1} e^{\frac{11 t}{18}} + c_{2} e^{\frac{11 t}{18}} t

Solution steps

324 y^{''} (t) - 396 y^{'} (t) + 121 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{11 t}{18}} + c_{2} e^{\frac{11 t}{18}} t

y = c_{1} e^{\frac{11 t}{18}} + c_{2} e^{\frac{11 t}{18}} t

9 y^{^{''}} + 6 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = \frac{- 13}{3}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0

8 y^{^{''}} - 25 y^{^{'}} + 3 y = 0

25 y^{^{''}} - 100 y^{^{'}} + 109 y = 0, y (0) = 9, y^{^{'}} (0) = 8

256 y^{^{''''}} = 81 y

What is the solution for 324y^{''}-396y^'+121y=0 ?
The solution for 324y^{''}-396y^'+121y=0 is y=c_{1}e^{(11t)/(18)}+c_{2}e^{(11t)/(18)}t