# Definition

**homogeneous**. If that equation equals $\cos(x)$ or $e^x$, there'll be no (easily solvable) answer.

# Method- Normal

Let's begin with an example: $$2x^2y" + xy'-15y = 0$$ We begin by guessing $x^r$ $$2x^2(r)(r-1)x^{r-2} + x(r)x^{r-1} - 15x^r = 0 \to 2r(r-1)x^r + rx^r - 15x^r = 0$$ We divide through by $x^r$ to get our roots. $$2r(r-1) + r - 15 = 0 \to 2r^2 - r + 15 = 0 \to (r-3)(2r + 5) = 0$$ We get our answer: $$y = c_1x^3 + c_2x^{-\frac{5}{2}}$$

# Repeated Roots

# Imaginary Roots

David Witten