All of the differential equatons we will see in this course can be written in the form $$ N[y] = f(t),$$ where $N$ is a differential operator. A differential operator is much like a function, except instead of inputting a number and getting another number, we input a function and get another function.

Example: The derivative operator $D[y] = \frac{d}{dt}y$ is the simplest differential operator. It simply takes a function $y$ and returns the derivative of $y$. For example if $y = t^2$, then $D[y] = 2t.$

Definition: A differential operator $L$ is linear if for any differentiable functions $y(t)$ and $z(t)$ and any $a\in\mathbb{R}$ we have

1. $L[a\cdot y] = a L[y]$
2. $L[y+z] = L[y] + L[z]$

Example: The derivative operator is linear since $$D[a y] = \frac{d}{dt} (a y) = a \frac{d}{dt} y = a D[y]$$ and $$D[y + z] = \frac{d}{dt}(y+z) = \frac{d}{dt}y +\frac{d}{dt}z = D[y]+D[z]$$

To show that a differential operator is not linear, we only need to find an example function or constant for which it does not satisfy the definition. The differential operator $$N[y] = \sqrt{y^2+\left(\frac{d}{dt}y\right)^2}$$ is not linear. Choosing $y = t$ and $z = -t$ gives $$N[y+z] = N[t-t] = N[0] = 0$$ but $$N[y] + N[z] = \sqrt{t^2 + 1} + \sqrt{(-t)^2+1} = 2\sqrt{t^2+1} \ne 0$$

First Order Linear ODEs

Any first order linear differential equation can be written as $$L[y] = f(t)$$ where $L[y] = \frac{dy}{dt}+q(t)y$ for some function $q(t)$. In particular, if $q(t) = 0$, then $L$ is just the derivative operator. In this case, we can find the solution $y(t)$ by just integrating both sides of the equation with respect to $t$.

Example: Find the solution to the initial value problem \begin{align} \frac{d}{dt}y &= \ln(t), && y(1) = 1. \end{align}

Solution: Integrating both sides gives \begin{align} \int \frac{dy}{dt} dt &= \int \ln(t) dt. && \text{by parts:}~~ u = \ln(t),~~v' = 1\\ \int dy &= t \ln (t) +\int \frac{1}{t}t dt\\ y + C_1 &= t\ln(t)+t + C_2 \end{align} We get arbitrary constants on both sides of the equatio from the integration. However, we can introduce a new constant $C= C_2-C_1,$ in order to write the general solution as $$ y(t) = t\ln(t)+t + C.$$ This is a one-parameter family of solutions. We need to find the particular solution that satisfies the initial condition $y(1)=1$ $$ 1 = y(1) = (1)\ln(1)+1 + C.$$ Solving for $C$ gives $C = 0.$ Therefore the solution to the initial value problem is $$y(t) = t\ln(t)+t$$

If $q(t)$ is not the zero function, we can use a similar strategy to write the equation in a way that we can integrate. The idea is to find a function $\mu(t)$ that we can multiply by both sides so that we can integrate directly. Multiplying by $\mu(t)$ gives $$\mu(t) \frac{dy}{dt} + \mu(t)q(t)y = \mu(t)f(t).$$ If we choose $\mu$ so that $\frac{d}{dt}\mu(t) = \mu(t)p(t)$, then the left hand side of this equation becomes $$ \mu(t) \frac{d}{dt}y(t) + y(t) \frac{d}{dt}\mu(t) = \frac{d}{dt} (\mu(t)y(t))$$ by the product rule. Then the ordinary differential equation becomes $$ \frac{d}{dt}(\mu(t) y(t)) = \mu(t)f(t),$$ which we can solve by integrating both sides. The function $\mu(t)$ is called an integrating factor, and can be found using the formula $$ \mu(t) = e^{\int q(t) dt}.$$

Example: Find a one-parameter family of solutions to $$\frac{dy}{dt}+\frac{1}{t}y = \frac{\ln(t)}{t^2}$$ for $t$ in $(0,\infty).$

Solution: An integrating factor is $\mu(t) = e^{\int 1/t dt} = e^{\ln(t)} = t.$ Then multiplying the ode through by $\mu(t) = t$ we get $$t \frac{dy}{dt} + y = \frac{\ln(t)}{t}.$$ By design the right hand side satisfies $$\frac{d}{dt}(ty) = \frac{\ln(t)}{t}$$ integrating gives \begin{align} \int \frac{d}{dt}(ty)dt &= \int \frac{\ln(t)}{t} dt && \text{by substitution:}~~u = \ln(t)\\ ty &= \int u du && \hspace{1.25in} du = \frac{1}{t}dt\\ y &= \frac{(\ln(t))^2}{2t}+ \frac{C}{t}. \end{align}

Exercises

• Consider the differential operator $$L[y] = t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt} + y.$$
  1. Show that $L$ is linear.
  2. Find $L[t^{-1}]$
  3. Find $L[\ln(t)]$
  4. Find $L[\cos(\ln(t))]$
• For each of the following odes, find a one parameter family of solutions.
  1. $\frac{dy}{dt} +\ln(t)y = 0$ for $t \in (0,\infty).$
  2. $t^2\frac{dy}{dt}+ty = t$ for $t\in (0,\infty).$
  3. $\frac{du}{dx} +2u = 1$ for $x\in(-\infty,\infty).$