What is the solution for 6y^{''}-8y^'+6y=0 ?

The solution for 6y^{''}-8y^'+6y=0 is y=e^{(2t)/3}(c_{1}cos((sqrt(5)t)/3)+c_{2}sin((sqrt(5)t)/3))

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6y^{''}-8y^'+6y=0

6 y^{^{''}} - 8 y^{^{'}} + 6 y = 0

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

Solution steps

6 y^{''} (t) - 8 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

What is the solution for 6y^{''}-8y^'+6y=0 ?
The solution for 6y^{''}-8y^'+6y=0 is y=e^{(2t)/3}(c_{1}cos((sqrt(5)t)/3)+c_{2}sin((sqrt(5)t)/3))