Find the general solution of the given second-order differential equation. \[ 2 y^{\prime \prime}-5 y^{\prime}+6 y=0 \] \( y(x)=\square \)

To find the general solution of the second-order differential equation: \[ 2 y'' - 5 y' + 6 y = 0 \] we start by finding the characteristic equation associated with the differential equation. Assume a solution of the form \( y = e^{rx} \). Plugging this into the differential equation gives: \[ 2r^2 e^{rx} - 5r e^{rx} + 6 e^{rx} = 0 \] Dividing through by \( e^{rx} \) (which is never zero): \[ 2r^2 - 5r + 6 = 0 \] This is a quadratic equation in \( r \). To solve it, we calculate the discriminant: \[ D = (-5)^2 - 4 \cdot 2 \cdot 6 = 25 - 48 = -23 \] Since the discriminant is negative (\( D < 0 \)), the roots are complex: \[ r = \frac{5 \pm \sqrt{-23}}{4} = \frac{5}{4} \pm \frac{\sqrt{23}}{4}i \] Letting \( \alpha = \frac{5}{4} \) and \( \beta = \frac{\sqrt{23}}{4} \), the general solution for the differential equation with complex roots is: \[ y(x) = e^{\alpha x} \left( C_1 \cos(\beta x) + C_2 \sin(\beta x) \right) \] Substituting the values of \( \alpha \) and \( \beta \): \[ y(x) = e^{\frac{5}{4}x} \left( C_1 \cos\left(\frac{\sqrt{23}}{4}x\right) + C_2 \sin\left(\frac{\sqrt{23}}{4}x\right) \right) \] where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. **Final Answer:** After simplifying, a general solution is e^(5⁄4 x) times (C₁ cos(√23⁄4 x) plus C₂ sin(√23⁄4 x)). Thus, y(x) = e^(5/4x)[C₁ cos(√23/4 x) + C₂ sin(√23/4 x)]