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. $$ y=c_{1} e^{-x}+c_{2} x e^{-x} $$

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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.

$$
y = c_{1} e^{-x} + c_{2} e^{3 x}
$$
The general solution is $$ y = c_{1} e^{-x} + c_{2} e^{3 x} $$.

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