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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} - y = 0

9 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

y^{^{''}} - a y^{^{'}} - b y = 0

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