Please help me with the following question:
Find a second order differential equation with constant coefficients for which the following function is a solution.
$3e^{2x} - 2e^{5x}$
Theorem:
Let $a_1$, $a_2$ be constants and consider the equation $L(y) = y'' + a_1y' + a_2y = 0$. If $r_1$, $r_2$ are distinct roots of the characteristic polynomial p, where $p(r) = r^2+a_1r+a_2$, then the functions $φ_1$, $φ_2$ are defined by $φ_1(x) = e^{r_1x}$ and $φ_2(x) = e^{r_2x}$ are solutions of $L(y) = 0$.
My attempt:
In this case, since $r_1 = 2$ and $r_2 = 5$, so the characteristic polynomial is $(r-2)(r-5) = r^2 - 7r + 10$. Therefore, the differential equation is $y''-7y' + 10y = 0$.
I am wondering how does the solution contain coefficients $3$ and $-2$? Please help!