# Ordinary Differential Equations Exercise 4.4 — A Fun Twist

I love when textbook exercises have little hidden twists to them. Surprises give character to the equations. Take Exercise 4.4 from Ordinary Differential Equations by Tenenbaum and Pollard:

Show that
$$y = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6})$$
is a 3-parameter family of solutions of the differential equation
$$y^{\prime\prime\prime} – 3y^{\prime\prime} + 3y’ – y – e^{x} = 0$$

You could proceed using the mechanical strategy of explicitly computing the third derivative of $y$ with respect to $x$ then solving for $c_1$, $c_2$ and $c_3$:

\begin{align} y & = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6}) \\ y’ & = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6}) + e^x(c_2 + 2c_3x + \frac{x^2}{2}) \\ y^{\prime\prime} & = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6}) + e^x(c_2 + 2c_3x + \frac{x^2}{2}) \\ & + e^x(c_2 + 2c_3x + \frac{x^2}{2}) + e^x(2c_3 + x) \\ y^{\prime\prime} & = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6}) + 2e^x(c_2 + 2c_3x + \frac{x^2}{2}) + e^x(2c_3 + x) \\ y^{\prime\prime\prime} & = e^x(c_1 + c_2x + c_3x^2 + \frac{x^3}{6}) + e^x(c_2 + 2c_3x + \frac{x^2}{2}) \\ & + 2e^x(c_2 + 2c_3x + \frac{x^2}{2}) + 2e^x(2c_3 + x) \\ & + e^x(2c_3 + x) + e^x \\ \end{align}

Yikes! We could certainly sit down and solve for $c_1$, $c_2$ and $c_3$, but this problem is already getting tedious and repetitive. One of the things that makes this particular textbook great is “tedious and repetitive” is a huge hint that we are missing something interesting. Time to start investigating.

Here’s something neat — $y$ is hiding in the equation for its first derivative:
$$y’ = y + e^x({c_2 + 2c_3x + \frac{x^2}{2}}) \\$$

Okay. Finding the second derivative from this equation is a bit simpler than the first time around:
$$y^{\prime\prime} = y’ + e^x({c_2 + 2c_3x + \frac{x^2}{2}}) + e^x(2c_3 + x) \\$$

Hey! We can rewrite $e^x({c_2 + 2c_3x + \frac{x^2}{2}})$ as $y' - y$, so we have a pretty simple second derivative:
\begin{align} y^{\prime\prime} & = y’ + (y’ – y) + e^x(2c_3 + x) \\ y^{\prime\prime} & = 2y’ – y + e^x(2c_3 + x) \end{align}

One more derivative to go:
$$y^{\prime\prime\prime} = 2y^{\prime\prime} – y’ + e^x(2c_3 + x) + e^x$$

Once more, we’ve already seen one of the terms in previous function. Substituting $y^{\prime\prime} - 2y' + y$ for $e^x(2c_3 + x)$ gives us
$$y^{\prime\prime\prime} = 3y^{\prime\prime} – 3y’ + y + e^x$$

From here the differential equation we were asked to show arises in a very natural and satisfying manner.

This exercise made me feel like I was on the right track while studying. I felt clever even though I think it’s clear that this is the path the authors intended. It was a great reminder to stay awake and observant while practicing. It’s important to start with “follow-your-nose” type exercises — ones that can be completed using mechanical processes like the first approach we tried — to verify you have a solid grasp on the fundamentals of a topic. However, those are just the first step. This text does a great job of giving the reader practice with the basics before subtly inviting the reader to explore on their own. It is a great recreational math book.