# How to explain to a high school student why a linear differential equation is linear?

My mother is teaching a high school course on multivariable calculus, and they were studying linear differential equations of the form $$y' + P(x) y = Q(x),$$ and the question of why this equation is called "linear" came up.

In terms that these students are familiar with, since they haven't been exposed to linear algebra yet, my thought was to say that the equation for a line, $y = mx+b$, is "linear" in $x$ (ignoring the technicality that it's actually an affine equation, not a linear one), because it's in the form "coefficient times $x$", and then we allow another term which is just a lonely coefficient. And then we extend this notion to saying that the above differential equation is "linear" in $y$ and $y'$, but this time the coefficients are allowed to be functions of $x$.

That's probably a good enough hand-wavy explanation to help students remember the definition, at the very least. I couldn't really think of a good reason why it "should", a priori, be kosher to allow coefficients to be functions of $x$ here. At that point it seems to me like you just have to get into the linear algebraic definition of linearity, which, being completely foreign to the students... it just seems to be a bit too deep of a rabbit hole for this purpose.

So my question is: does anyone have a better way of approaching this? And if you think my hand-wavy explanation above is largely acceptable, is there a way you can explain why multiplying by non-constant functions of $x$ "should" still be considered linear in $y$?

• What high school teaches this stuff...? Because I think I want to send my children there. – IAmNoOne Sep 28 '14 at 3:00
• Herb Gross explains it well in this video. – IAmNoOne Sep 28 '14 at 3:03
• Have her give them an example of a differential equation that is nonlinear, so they can appreciate why it is far harder to create new solutions from old solutions in the nonlinear case but not so hard in the linear case. – KCd Sep 28 '14 at 3:21
• @Nameless Come to San Diego! It's standard (at a few schools) here to teach through Calculus III and Linear Algebra by Senior year, and with growing frequency Junior year. – Aza Sep 28 '14 at 5:23
• @Nameless This school is a private school, but it's not unheard of for larger public schools/magnet schools to have more advanced mathematics courses. – Dustan Levenstein Sep 28 '14 at 5:57

The differential equation $y'+Py=Q$ is linear because the underlying homogeneous problem $y'+Py=0$ satisfies the linearity property that when $y_1, y_2$ are solutions the arbitrary linear combination $c_1y_1+c_2y_2$ is once more a solution. It is linearity of the solution set rather than the explicit form of the differential equation itself which is of interest. Of course, there is also the superposition principle, if we replace $Q$ with $c_1Q_1+c_2Q_2$ then the solutions of $y'+Py=Q_1$ and $y'+Py=Q_2$ superpose to give solutions of $y'+Py=c_1Q_1+c_2Q_2$ hence the net-cause is a sum of the individual effects. All of these features are characteristic of linear systems.

As Nameless pointed out in his comment, talking about what does Gross' "$L$-machine" do to linear combinations of functions can be useful and easy to understand, given that students already know derivatives properties:

$$L(y) = y'+P(x)y,$$

$$L(ay_1+by_2) = aL(y_1)+bL(y_2).$$

The derivative machine ($y'$) is already known to be "linear". We are expanding this machine carefully (+$P(x)y$) in a way in which we don't lose that "linearity".

I think the best way might be to think about what would happen if we had two solutions to this $y_1$ and $y_2$. Then, we could consider that, for $a+b=1$, the function $ay_1+by_2$ is also a solution. It's rather obvious that this is a line if they happen to know linear algebra, but otherwise, here's how I'd explain it:

Draw the plane with axes $a$ and $b$ and the line $a+b=1$ - which the students should be able to recognize as a line given what is taught in high school. Notice that this passes through the points $(1,0)$ and $(0,1$). Consider that, if $(a,b)=(1,0)$, then $ay_1+by_2=y_1$ - so the point $(1,0)$ on the coordinate plane represents the function $y_1$. Similarly, the point $(0,1)$ represents the function $y_2$. Then, as an example, the middle point $\left(\frac{1}2,\frac{1}2\right)$ can be viewed as the average $\frac{y_1+y_2}2$, which should be another familiar form. You could also look at other points as weighted averages. Doing this should give them a sense that this line in the $ab$ coordinate plane represents that any function in some sense "between" or arising from an average of two others must also be a solution - and the students, as a bonus, have the visual of a line on a plane, which, at the very least, acts as a mnemonic if not to give insight.

Just tell them that the word "linear" (when used outside of elementary algebra) is defined to mean "homogeneous of degree 1" and "additive". Give an example of each of those 2 properties, explain that it's really the homogeneous equation that's linear, and move on. It's just a definition, and they'll get a better idea of it when they get to linear algebra so there's no reason to waste time on it.

/rant Although, I should say that taking multivariable or ODEs without first having had linear algebra is not a great idea. I wish high schools would quit doing this. The ideas of multivariable calculus and ODEs can sink in a LOT deeper, if one isn't starting from scratch. And because of the way AP credit works, the students may be moving on to even higher level math courses (PDEs, etc) immediately after they get to college without ever really understanding what they were doing with all those Jacobians, tangent spaces, and Wronskians. /endrant

• That's fair enough if you have students who are willing to accept a definition without questioning it. However, I personally think students who have advanced to this level should be encouraged to ask questions, rather than being discouraged from asking questions. – Dustan Levenstein Sep 28 '14 at 6:06
• @DustanLevenstein I wasn't suggesting that questions should be discouraged, but most students would be fine with a teacher saying "it's just a definition". And for the couple of students actually interested, the teacher could tell them to come back her office after school and she'll explain a little linear algebra to them -- that way it doesn't interrupt class with things that most students in the class won't care about and which won't affect their grades. – user179068 Sep 28 '14 at 15:21
• Fair enough. I still think there are better ways to approach this particular question, but to each their own. – Dustan Levenstein Sep 28 '14 at 19:27