What is the solution for 4y^{''}-4y^'+13y=0 ?

The solution for 4y^{''}-4y^'+13y=0 is y=e^{t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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

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

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

What is the solution for 4y^{''}-4y^'+13y=0 ?
The solution for 4y^{''}-4y^'+13y=0 is y=e^{t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))