What is the solution for 4y^{''}-4y^'+17y=0 ?

The solution for 4y^{''}-4y^'+17y=0 is y=e^{t/2}(c_{1}cos(2t)+c_{2}sin(2t))

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4y^{''}-4y^'+17y=0

4 y^{^{''}} - 4 y^{^{'}} + 17 y = 0

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

4 y^{^{''}} - 2 y^{^{'}} + 9 y = 0

y^{^{''}} + 19 y^{^{'}} + 90 y = 0

y^{^{'''}} - 4 y^{^{''}} - 41 y^{^{'}} - 36 y = 0

y^{^{'''}} + 10 y^{^{''}} - y^{^{'}} - 3 y = 0

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 12

What is the solution for 4y^{''}-4y^'+17y=0 ?
The solution for 4y^{''}-4y^'+17y=0 is y=e^{t/2}(c_{1}cos(2t)+c_{2}sin(2t))