What is the solution for 2y^{''}-5y^'+6y=0 ?

The solution for 2y^{''}-5y^'+6y=0 is y=e^{(5t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))

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2y^{''}-5y^'+6y=0

2 y^{^{''}} - 5 y^{^{'}} + 6 y = 0

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

Solution steps

2 y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

What is the solution for 2y^{''}-5y^'+6y=0 ?
The solution for 2y^{''}-5y^'+6y=0 is y=e^{(5t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))