What is the solution for 4y^{''}-8y^'+5y=0 ?

The solution for 4y^{''}-8y^'+5y=0 is y=e^t(c_{1}cos(t/2)+c_{2}sin(t/2))

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4y^{''}-8y^'+5y=0

4 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

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

Solution steps

4 y^{''} (t) - 8 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 4 y = 0, y (\frac{π}{6}) = 5, y (0) = - 2

y^{^{''''}} - 14 y^{^{'''}} + 49 y^{^{''}} = 0

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

y^{^{''}} - 12 y^{^{'}} + 45 y = 0

What is the solution for 4y^{''}-8y^'+5y=0 ?
The solution for 4y^{''}-8y^'+5y=0 is y=e^t(c_{1}cos(t/2)+c_{2}sin(t/2))