What is the solution for 36y^{''}-36y^'+13y=0 ?

The solution for 36y^{''}-36y^'+13y=0 is y=e^{t/2}(c_{1}cos(t/3)+c_{2}sin(t/3))

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36y^{''}-36y^'+13y=0

36 y^{^{''}} - 36 y^{^{'}} + 13 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

Solution steps

36 y^{''} (t) - 36 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

y^{^{''}} - 2 y^{^{'}} - y^{^{'}} + 2 y = 0

y^{^{''}} + 24 y^{^{'}} - 10 y = 0

81 y^{^{''}} - 64 y = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

y^{^{'''}} - 3 y^{^{''}} - 46 y^{^{'}} - 72 y = 0

What is the solution for 36y^{''}-36y^'+13y=0 ?
The solution for 36y^{''}-36y^'+13y=0 is y=e^{t/2}(c_{1}cos(t/3)+c_{2}sin(t/3))