Recall that a differential equation can always be written as $$L[y] = f(t)$$ where $L$ is a differential operator and $f$ is some function of $t$. Differential equations are linear if $L$ is a linear differential operator. That is, $L$ satisfies

1. $L[ay] = a L[y]$ for any constant $a\in\mathbb{R}$
2. $L[y_1+y_2] = L[y_1] +L[y_2]$ for any two functions $y_1$ and $y_2$

In general, this means thatany $n$th order linear differential equation can be written in the form $$ c_n(t)\frac{d^ny}{dt^n}+...+c_1(t)\frac{dy}{dt}+c_0(t)y = f(t).$$ The functions $c_n(t)$ are called the coefficients of the linear operator. We will spend most of our time looking at differential equations where the coefficients are constants.

Definition: An ODE is called homogeneous if $f(t)=0$ and inhomogeneous if $f(t) \ne 0.$

We'll start by looking for solutions to homogeneous differential equations, and show that they can be used to help find solutions to non-homogenous equations.

Example: Consider the second order homogeneous differential equation $$\frac{d^2y}{dt^2} - y = 0.$$ We can rewrite this as $$\frac{d^2y}{dt^2}=y,$$ and so solving this equation is the same as asking to find a function that is its own second derivative.

The linear superposition principle: If $L$ is a linear differential operator, and $y_1$ and $y_2$ are both solutions to $L[y]=0$, then any linear combination of $y_1$ and $y_2$ are also solutions.

As a consequence of this, $y_0 = 0$ is always a solution to homogeneous linear differential equations. We call this the trivial solution. We are often only interested in other solutions, so we will be looking for nontrivial solutions.

Practice: Find all of the solutions to $$\frac{d^2y}{dt^2} +y = 0$$

Theorem: Every linear $n$th order differential equation with constant coefficients has at least one solution of the form $y = e^{\lambda t}$ for some $\lambda \in \mathbb{C}.$ We can see this by writing \begin{align} L[e^{\lambda t}] &= c_n \frac{d^n e^{\lambda t}}{dt^n} +...+ c_1\frac{d e^{\lambda t}}{dt}+c_0 e^{\lambda t}\\ & =c_n \lambda ^n e^{\lambda t} + ... + c_1 \lambda e^{\lambda t} + c_0 e^{\lambda t}\\ &=e^{\lambda t} ( c_n \lambda ^n + ... + c_1 \lambda + c_0) \end{align} This is a polynomial in $\lambda$, and by the fundamental theorem of algebra, every non-constant polynomial has at least one complex root.

Example: Find a non-trivial solution to the differential equation $$ \frac{d^2 y}{dt^2} + 5 \frac{dy}{dt}+ 6 y = 0.$$

Solution: By the previous theorem, we know there is at least one solution of the form $y = e^{\lambda t}.$ Subbing this in gives \begin{align} \frac{d^2}{dt^2}e^{\lambda t} + 5 \frac{d}{dt} e^{\lambda t} + 6 e^{\lambda t} &= 0\\ e^{\lambda t} ( \lambda^2 + 5 \lambda + 6) &= 0, \end{align} which is zero if $\lambda = -2$ or if $\lambda = -3.$ Therefore one solution to the problem is $y_1(t) = e^{-2t}$ and another solution is $e^{-3t}.$ We can combine these two solutions to get the two-parameter family of solutions $$y = C_1 e^{-2t}+C_2 e^{-3t}.$$

Practice Problems

Find at least one non-trivial solution to each of the following differential equations

1. $$ 4\frac{d^2y}{dt^2} +2\frac{dy}{dt} -6 = 0$$
2. $$ 9\frac{d^2y}{dt^2} - 25 y = 0$$
3. $$ \frac{d^2y}{dt^2}+4\frac{dy}{dt}+4 y = 0$$
4. $$ \frac{d^2y}{dt^2}+7\frac{dy}{dt} = 0$$