What is the solution for 2y^{''}-3y^'+5y=0 ?

The solution for 2y^{''}-3y^'+5y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(31)t)/4)+c_{2}sin((sqrt(31)t)/4))

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2y^{''}-3y^'+5y=0

2 y^{^{''}} - 3 y^{^{'}} + 5 y = 0

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 0, y (0) = 2, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} = 0, y (0) = cos (x)

x^{^{''}} + 6 x^{^{'}} + 25 x = 0

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

y^{^{''}} - 12 y^{^{'}} - 28 y = 0

What is the solution for 2y^{''}-3y^'+5y=0 ?
The solution for 2y^{''}-3y^'+5y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(31)t)/4)+c_{2}sin((sqrt(31)t)/4))