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Let's say you have a homogeneous (equal to 0) differential equation $$Ay'' + By' + Cy = 0$$
Notice that because the only coefficients are constants, then y, y', and y'' all must be the same general form. What do we know that gives pretty much the same thing back when we differentiate it? $e^x$. We don't want to guess e^x, because $e^{2x}$could also work, so we guess $e^{rx}$.

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$$y'' + 3y' - 10y = 0$$
Guess $e^{rx}$
$$r^2(e^{rx}) + 3r(e^{rx}) - 10(e^{rx}) = (r^2 + 3r - 10)(e^{rx}) = 0$$ Because $e^x$ is never 0, we can divide through by that to get r = -2, 5.
Therefore, our solutions are $e^{-2x}, e^{5x}$.

Here is an important thing to consider. If $L(y) = Ay'' + By' + Cy$, that means it's a linear function, because nothing's squared. Because it's linear, we can say $L(y_1 + y_2) = L(y_1) + L(y_2)$. Also $L(cy_1) = cL(y_1)$. By this same principle if $L(y_1) = 0$ and $L(y_2) = 0$, then $L(c_1y_1 + c_2y_2) = 0$.
Therefore, our final solution is
$$\boxed{c_1e^{-2x} + c_2e^{5x}}$$

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Let's say you have this:
$$y'' + y = 0$$
You turn this into:
$$(r^2 + 1)e^{rx} = 0$$
$$r = \pm 1 \to e^{ix}, e^{-ix}$$
$$e^{ix} = \cos(x) + i\sin(x)$$
$$e^{-ix} = \cos(x) - \sin(x)$$
$$c_1(\cos(x) + i\sin(x)) + c_2(\cos(x) - i\sin(x))$$
$$(c_1 + c_2)\cos(x) + (c_1 - c_2)i\sin(x)$$
Side note: if $a + bi = 0$, then $a,b = 0$, because 1 and $i$ are linearly independent.
So, the solutions are
$$d_1\cos(x) + d_2\sin(x)$$

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Let's look at this:
$$y'' + 4y' + 4y = 0$$
Then, $(r+2)^2$. We know we need two roots, but we only have one r. What do we do? We have $e^{-2x}$ twice. For the sake of simplicity, I have omitted the proof, but all we have to do is add x to one solution, so we get. For a thrice repeated root we add x and $x^2$.
$$\boxed{c_1e^{-2x} + c_2xe^{-2x}}$$