Hello joe,
We are given the 2nd order homogeneous ODE:
$$y''-2y'-15y=0$$
Part 1.) Letting:
$$y=e^{mx}$$
We can see that:
$$\frac{d^ny}{dx^n}=m^ne^{mx}$$ where $n\in\mathbb{N}$
Hence, substituting this function into the ODE, the ODE becomes:
$$m^2e^{mx}-2me^{mx}-15e^{mx}=0$$
Divide through by $$e^{mx}\ne0$$ and we have:
$$m^2-2m-15=0$$
This is what is referred to as the characteristic or auxiliary equation. Factoring, we obtain:
$$(m-5)(m+3)=0$$
And so we find the characteristic roots are:
$$m=-3,\,5$$
These are the two values of $m$ for which $y=e^{mx}$ is a solution to the given ODE.
Part 2) Let:
$$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$
As before, given the linearity of differentiation, we can see:
$$y^{(n)}(x)=c_1m_1^ne^{m_1x}+c_2m_2^ne^{m_2x}$$
Hence, substituting this function into the ODE, we obtain:
$$\left(c_1m_1^2e^{m_1x}+c_2m_2^2e^{m_2x} \right)-2\left(c_1m_1e^{m_1x}+c_2m_2e^{m_2x} \right)-15\left(c_1e^{m_1x}+c_2e^{m_2x} \right)=0$$
Arrange on like terms:
$$\left(c_1m_1^2e^{m_1x}-2c_1m_1e^{m_1x}-15c_1e^{m_1x} \right)+\left(c_2m_2^2e^{m_2x}-2c_2m_2e^{m_2x}-15c_2e^{m_2x} \right)=0$$
Factor both expressions:
$$c_1e^{m_1x}\left(m_1^2-2m_1-15 \right)+c_2e^{m_2x}\left(m_2^2-2m_2-15 \right)=0$$
Now, since we have defined $m_1$ and $m_2$ wot be the two roots of the quadratic characteristic equation $m^2-2m-15=0$, we must therefore have:
$$c_1e^{m_1x}\cdot0+c_2e^{m_2x}\cdot0=0$$
$$0=0$$
This shows that the function:
$$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$
is the general solution to the given ODE.