Solve the following differential equation $$ y^{\prime \prime}+2 y^{\prime}-8 y=0 $$ Select one: a. $$ y=c_{1} e^{-3 x}+c_{2} e^{2 x} $$ b. $$ y=c_{1} e^{x}+c_{2} e^{-8 x} $$ c. $$ y=c_{1} e^{2 x}+c_{2} e^{-4 x} $$ d. $$ y=c_{1} e^{x}+c_{2} e^{-2 x} $$ e. $$ y=c_{1} e^{-2 x}+c_{2} e^{4 x} $$

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We start by writing the differential equation:

$$
y^{\prime \prime} + 2y^{\prime} - 8y = 0
$$

The next step is to form the characteristic equation by assuming a solution of the form $$ y = e^{rx} $$. Plugging this into the differential equation gives:

$$
r^2 e^{rx} + 2r e^{rx} - 8e^{rx} = 0
$$

Since $$ e^{rx} \neq 0 $$, we can divide through by $$ e^{rx} $$ to obtain the characteristic equation:

$$
r^2 + 2r - 8 = 0
$$

Now, we solve the quadratic equation:

$$
r^2 + 2r - 8 = 0
$$

Using the quadratic formula:

$$
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = 2 $$, and $$ c = -8 $$. Substituting these values in:

$$
r = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}
$$

Simplify further:

$$
r = \frac{-2 \pm 6}{2}
$$

This gives:

$$
r = \frac{4}{2} = 2 \quad \text{or} \quad r = \frac{-8}{2} = -4
$$

Thus, the general solution for the differential equation is:

$$
y = c_{1} e^{2x} + c_{2} e^{-4x}
$$

Comparing with the given options, the correct answer is:

c. $$ y = c_{1} e^{2x} + c_{2} e^{-4x} $$
The solution is $$ y = c_{1} e^{2x} + c_{2} e^{-4x} $$, which corresponds to option c.

Solve the following differential equation

Select one:
a.
b.
c.
d.
e.

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