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 with respect to then solving for , and :

$$

\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 , and , 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 — 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 as , 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 for 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.