Associated general solution of C.F.
Complementary functions are homogeneous
$$ \begin{align*} (a_{0}D^{n}+a_{1}D^{n-1}+…+a_{n})y=0\\ \\ \\ \\ -&(1)\\ (D-m_{1})(D-m_{2})…(D-m_{n})=0\\ \\ \\ \\ -&(2)\\ \end{align*} $$
Solutions of any one of $(D-m_k)y=0$ is also a solution of (2). Can be proven by swapping the terms, because that changes nothing:
$$ \begin{align*} (D-m_k)y=0\\ (D-m_1)(D-m_2)…(D-m_n)…\\ \\ (D&-m_k)y=0\\ (D-m_1)(D-m_2)…(D-m_n)…\\ \\ &0=0\\ 0==0 \end{align*} $$
Therefore, it’s a solution of the original ODE as well.
Trial solution #
$$ \begin{align*} (D-m_{k})y=0\\ \dd{y}{x}-m_{k}y=0\\ y=C_{k}e^{m_{k}x} \end{align*} $$
Substituting into the main to check #
$$ \begin{align*} (a_{0}D^{n}+a_{1}D^{n-1}+…a_{n})y=0\\ (a_{0}D^{n}+a_{1}D^{n-1}+…a_{n})e^{m_{k}x}=0\\ (a_{0}m^{n}+a_{1}m^{n-1}+…+a_n)e^{m_{k}x}=0 \end{align*} $$
Since the exponential term cannot get 0, the first term has to be=0. This term is called the Auxilary Equation
Can be found by replacing D’s by m’s. Auxilary equation of f(D)y=0 is: f(m)=0
Number of Roots of this equation = n, and they give us the n m’s where $y_k=e^{m_{k}x};k={1,2,…,n}$ satisfy the main ODE.
CASE I: Distinct roots of m #
If all m’s are distinct: $$ \begin{align*}\\ y_{1}=e^{m_{1}x},\\ y_{2}=e^{m_{2}x},\\ …\\ y_{n}=e^{m_{n}x} \end{align*} $$
are linearly independent and general solution of f(D)y=0 is:
$$ y_c(x)=c_{1}e^{m_{1}x}+c_{2}e^{m_{2}x}+c_{3}e^{m_{3}x}+…+c_{n}e^{m_{n}x} $$
Utilizing Linear Higher Order Differential Equations
CASE II: Equal roots of m #
If for a ‘k’, mâ‚– is repeating i times, then it’s contribution in the linear combination would be:
$$ (C_1+C_2x+…C_ix^{i-1})e^{m_{k}x} $$
If m is complex #
$$ \begin{align*} \text{two roots associated with }e^{m_{k}x}:\\ e^{x(a+bi)}e^{x(a-bi)}=e^{ax}(c_1\cos(bx)+c_2\sin(bx)) \end{align*} $$
If they’re repeating, $c_1,c_2$ will become $(c_1+c_2x+…),(c_3+c_4x+…)$