Homogeneous linear differential equation with constant coefficients

Definition

A homogeneous linear differential equation with constant coefficients, which can also be thought of as a linear differential equation that is simultaneously an autonomous differential equation, is a differential equation of the form:

$y^{(k)}+p_{k-1}y^{(k-1)}+\dots +p_{1}y'+p_{0}y=0$

where $p_{0},p_{1},\dots ,p_{k-1}$ are all constants (i.e., real numbers). Here $y$ is the dependent variable and $x$ (which does not appear explicitly above) is the independent variable with respect to which the differentiations occur.

Solution method

Consider the following polynomial:

$t^{k}+p_{k-1}t^{k-1}+\dots +p_{1}t+p_{0}$

This polynomial is called the characteristic polynomial of the differential equation. We consider various cases:

Case	Solution in that case
The polynomial has pairwise distinct real roots $\alpha _{1},\alpha _{2},\dots ,\alpha _{k}$	The solution space has basis $e^{\alpha _{1}x},e^{\alpha _{2}x},\dots ,e^{\alpha _{k}x}$. In other words, the general solution is $C_{1}e^{\alpha _{1}x}+C_{2}e^{\alpha _{2}x}+\dots +C_{k}e^{\alpha _{k}x}$ where $C_{1},C_{2},\dots ,C_{k}$ are freely varying real parameters.
The polynomial splits completely into linear factors over the reals, but with possible repetitions. $\alpha _{1}$ occurs $s_{1}$ times, $\alpha _{2}$ occurs $s_{2}$ times, and so on till $\alpha _{r}$, which occurs $s_{r}$ times. We have $s_{1}+s_{2}+\dots +s_{r}=k$.	The solution space has basis all functions of the form $x^{d}e^{\alpha _{i}x}$ where $0\leq d<s_{i}$ with $d$ an integer. Thus, for each $i$, there are $s_{i}$ basis vectors corresponding to $\alpha _{i}$. We get a total of $k$ basis vectors.
The polynomial splits completely over the complex numbers into distinct linear factors, but some of the roots are not real	For any real root $\alpha$, use $e^{\alpha x}$ as a basis vector. Non-real roots occur in complex conjugate pairs. For a pair $a\pm ib$, choose the vectors $e^{ax}\cos(bx)$ and $e^{ax}\sin(bx)$. Combining, we get a basis of $k$ vectors.
The general case	For a real root $\alpha$ of multiplicity $s$, the $s$ basis vectors are $x^{d}e^{\alpha x},0\leq d<s$. For a pair of complex conjugates $a\pm ib$ of multiplicity $s$, the $2s$ basis vectors are $x^{d}e^{ax}\cos(bx),0\leq d<s$ and $x^{d}e^{ax}\sin(bx),0\leq d<s$.

Examples

Examples with distinct real roots

Consider the differential equation:

$y''-3y'+2y=0$

The characteristic polynomial is:

$t^{2}-3t+2$

We want to find the roots of this polynomial. The roots are $\alpha _{1}=1,\alpha _{2}=2$. Thus, the general solution is:

$y=C_{1}e^{x}+C_{2}e^{2x},\qquad C_{1},C_{2}\in \mathbb {R}$

Here is another example:

$y'''+2y''-2y'-4y=0$

The characteristic polynomial is:

$t^{3}+2t^{2}-2t-4$

The roots are $\alpha _{1}=-2,\alpha _{2}=-{\sqrt {2}},\alpha _{3}={\sqrt {2}}$. The general solution is thus:

$y=C_{1}e^{-2x}+C_{2}e^{-{\sqrt {2}}x}+C_{3}e^{{\sqrt {2}}x},\qquad C_{1},C_{2},C_{3}\in \mathbb {R}$

Examples with repeated real roots

Consider the differential equation:

$y''=6y'-9y$

First, we bring everything to one side, to get:

$y''-6y'+9y=0$

The characteristic polynomial is:

$t^{2}-6t+9$

This factors as $(t-3)^{2}$, so it has a root 3 with multiplicity 2. The general solution is thus:

$y=(C_{1}+C_{2}x)e^{3x},\qquad C_{1},C_{2}\in \mathbb {R}$

Examples with real and complex roots without repetitions

Consider the differential equation:

$y'''=y$

First, we bring everything to one side, to get:

$y'''-y=0$

The characteristic polynomial is:

$t^{3}-1$

The only real root is 1, corresponding to a solution basis element $e^{x}$. The complex roots are the roots of $t^{2}+t+1$, and they are complex conjugates:

${\frac {-1\pm {\sqrt {3}}i}{2}}$

The basis solutions for this pair are thus $e^{-x/2}\cos({\sqrt {3}}x/2)$ and $e^{-x/2}\sin({\sqrt {3}}x/2)$. The general solution is thus:

$y=C_{1}e^{x}+C_{2}e^{-x/2}\cos({\sqrt {3}}x/2)+C_{3}e^{-x/2}\sin({\sqrt {3}}x/2)$