What is the solution for 2y^{''}-3y^'+8y=0 ?

The solution for 2y^{''}-3y^'+8y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(55)t)/4)+c_{2}sin((sqrt(55)t)/4))

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2y^{''}-3y^'+8y=0

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

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{55 t}{4}) + c_{2} sin (\frac{55 t}{4}))

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{55 t}{4}) + c_{2} sin (\frac{55 t}{4}))

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{55 t}{4}) + c_{2} sin (\frac{55 t}{4}))

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (0) = - 1, y^{^{'}} (0) = 1

y^{^{''''}} - 3 y^{^{'''}} - 4 y^{^{''}} = 0

y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 2, y (2) = 1

2 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

What is the solution for 2y^{''}-3y^'+8y=0 ?
The solution for 2y^{''}-3y^'+8y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(55)t)/4)+c_{2}sin((sqrt(55)t)/4))