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. $L[ay]=aL[y]$ 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 $$c_n(t)\frac{d^{n}y}{d{t}^n}+...+c_1(t)\frac{dy}{dt}+c_0(t)y=f(t).$$ 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 $$\frac{d^{2}y}{d{t}^2}-y=0.$$ We can rewrite this as $$\frac{d^{2}y}{d{t}^2}=y,$$ 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 $$\frac{d^{2}y}{d{t}^2}+y=0$$

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 $$\begin{array}{rl}L[e^{\lambda t}]& =c_n\frac{d^n{e}^{\lambda t}}{d{t}^n}+...+c_1\frac{d{e}^{\lambda t}}{dt}+c_0{e}^{\lambda t}\\ & =c_n{\lambda }^n{e}^{\lambda t}+...+c_1\lambda e^{\lambda t}+c_0{e}^{\lambda t}\\ & =e^{\lambda t}(c_n{\lambda }^n+...+c_1\lambda +c_0)\end{array}$$ 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 $$\frac{d^{2}y}{d{t}^2}+5\frac{dy}{dt}+6y=0.$$

Solution: By the previous theorem, we know there is at least one solution of the form $y=e^{\lambda t}.$ Subbing this in gives $$\begin{array}{rl}\frac{d^2}{d{t}^2}{e}^{\lambda t}+5\frac{d}{dt}{e}^{\lambda t}+6e^{\lambda t}& =0\\ e^{\lambda t}({\lambda }^2+5\lambda +6)& =0,\end{array}$$ 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. $$4\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}-6=0$$
2. $$9\frac{d^{2}y}{d{t}^2}-25y=0$$
3. $$\frac{d^{2}y}{d{t}^2}+4\frac{dy}{dt}+4y=0$$
4. $$\frac{d^{2}y}{d{t}^2}+7\frac{dy}{dt}=0$$