Find the general solution of the given equation. \( y^{\prime \prime}-2 y^{\prime}-3 y=0 \) Choose the correct solution below. \( c_{1} \) and \( c_{2} \) are arbitrary constants. A. \( y=c_{1} e^{x}+c_{2} e^{-3 x} \) B. \( y=c_{1} \cos (-x)+c_{2} \sin (3 x) \) C. \( y=c_{1} e^{-x}+c_{2} e^{3 x} \) D. \( y=c_{1} e^{-x}+c_{2} x e^{-x} \)

To find the general solution of the differential equation: \[ y'' - 2y' - 3y = 0 \] we start by solving the characteristic equation associated with the differential equation: \[ r^2 - 2r - 3 = 0 \] Solving this quadratic equation: \[ r = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \] This gives two distinct real roots: \[ r_1 = 3 \quad \text{and} \quad r_2 = -1 \] Since the roots are distinct and real, the general solution of the differential equation is: \[ y = c_1 e^{3x} + c_2 e^{-x} \] Examining the given options: - **Option C**: \( y = c_1 e^{-x} + c_2 e^{3x} \) This matches our derived general solution, just with the terms swapped, which is perfectly acceptable since \( c_1 \) and \( c_2 \) are arbitrary constants. **Answer:** Option C. \[ y = c_{1} e^{-x} + c_{2} e^{3 x} \]