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Solve the initial value problem: $y'+2y=e^{-x}$ with $y(0)=4$

Our teacher gave me a mistake / unclear mark at the part which I will mark red here. But I don't understand why and I no longer have the opportunity to ask him.

This is an extract of my solution (it leads to the correct solution):

$$e^{2x} \cdot \frac{dy}{dx} + e^{2x} \cdot 2y = e^{2x} \cdot e^{-x} \Leftrightarrow$$

$$\Leftrightarrow \color{red}{\frac{d}{dx} \cdot e^{2x} \cdot y} = e^{x} \Leftrightarrow$$

$$\int{\frac{d}{dx} \cdot e^{2x} \cdot y} \text{ } dx = \int{e^x} \text{ } dx$$

$$...$$

I don't understand, is an explanation here really required? So I took that $y$ from $\frac{dy}{dx}$ and wrote it at the end of the term and now we got $\frac{d}{dx}$ at the front. $\frac{d}{dx}$ means "the derivative of.." (whatever comes after). And taking that derivative, we will get what we had before (the line before the red marked line), so they are equal to each other.

But how would you explain that (on paper) in a reasonable way? I don't know if my current explanation is fine.

Edit: When I handed this in, I used brackets too. The reason I got point deduction is "how do you get from here to there?" (first line to red line).

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    $\begingroup$ My recommendation here as in many other cases is not to use symbols like $\iff$ to connect the steps of a derivation. Write complete sentences in English (or the language in which you are instructed) with formulas embedded in them when necessary (but not merely as shorthand for words like "if" or "for all"). Tanner Swett's answer is similar to the style I would use. Of course it's also important to give an appropriate level of detail, as explained in Matthew Leingang's answer. $\endgroup$ – David K Oct 3 '17 at 17:40
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    $\begingroup$ If I were grading this, I would deduct points for exactly what is written in red. If you had written "$\frac{d}{dx}\{e^{2x}\cdot y\}=e^x$", I would give full credit. The notation, in particular (1) using "$\cdot$" after "$\frac{d}{dx}$" and (2) omitting grouping symbols around "$e^{2x}\cdot y$", is poor practice and should be corrected. $\endgroup$ – MPW Oct 3 '17 at 18:49
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I can't decide whether to offer my opinion (as a teacher) as an answer, or to vote to close because it's impossible to know your teacher's rationale. I guess the former.

There is a little bit of ambiguity in writing $\frac{d}{dx}\cdot e^{2x} \cdot y$ on the left-hand side of your highlighted equation. We don't normally multiply ($\cdot$) by the operator $\frac{d}{dx}$. I think you meant to say that the derivative is applied to the product $e^{2x} \cdot y$. I would prefer that be written as $\frac{d}{dx}\left(e^{2x}\cdot y\right)$.

Second point, and not having to do with your notation (which, as you say in your edit, was not as it was in the original submission). If this problem is assigned at a time when you are just learning how to solve linear ODEs with the integrating factor, it might also be the teacher wants you to show a few more steps. Once you move on to other new topics, this technique becomes routine, and steps can be skipped. But in some sense, you need to show that you can fill in the steps before you're afforded the privilege of skipping the steps.

I admit, it's possible that these reasons have nothing to do with the deduction. It could also be a straight-up mistake on the part of the instructor. But I usually ask for students to write up clear, explained answers, and this response lacks clarity in the notation and could possibly (depending on context) fall short of the expectations for explanation. Keep in mind that getting the right answer is not the only learning objective your teacher has for you. Being able to communicate mathematically, use symbols correctly, and justify steps, is also part of the learning.

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  • $\begingroup$ I had no idea that one question on one line of a test or off homework could engender such nuanced thinking solely on the methods of teaching. You must be a very gifted teacher. $\endgroup$ – Senex Ægypti Parvi Oct 3 '17 at 13:50
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    $\begingroup$ @SenexÆgyptiParvi that's very kind of you to say. Let's call it experience rather than talent. :-) $\endgroup$ – Matthew Leingang Oct 3 '17 at 18:19
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    $\begingroup$ @SenexÆgyptiParvi This is the norm, or if not, should be where it comes to teaching. . . $\endgroup$ – iheanyi Oct 3 '17 at 18:40
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    $\begingroup$ In my experience teaching and grading assignments, I've had thoughts along these lines very often. Like, on the order of half of all assignments submitted would have at least one point which makes me think like this. So I'd echo the position that it is (or should be) very normal. Not to say Matthew isn't a gifted teacher, but I'd say it takes a lot more than this to consider someone gifted. $\endgroup$ – David Z Oct 3 '17 at 23:15
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It took me a minute or two to "decipher" how you'd gotten from the first line to the second line, especially since I wasn't sure whether you were applying an operation to both sides of the equation, or just manipulating the two sides independently.

For an answer written in "quick" style, I think this would be a fine way to write this manipulation:

$$e^{2x} \cdot \frac{dy}{dx} + e^{2x} \cdot 2y = e^{2x} \cdot e^{-x}$$

Rearrange LHS and simplify RHS:

$$e^{2x} \cdot \frac{dy}{dx} + 2 e^{2x} \cdot y = e^x$$

Apply chain rule and product rule in reverse:

$$e^{2x} \cdot \frac{dy}{dx} + \frac{d}{dx} (e^{2x}) \cdot y = e^x$$ $$\frac{d}{dx} (e^{2x} \cdot y) = e^x$$

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  • $\begingroup$ As written, this omits relevant information from the original, specifically that all of the steps are reversible. $\endgroup$ – Hurkyl Oct 3 '17 at 20:49
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First of all, some parentheses would be nice. It is not clear if you mean

$$\frac{d}{dx}\left(e^{2x}\right)\cdot y$$

or $$\frac{d}{dx}\left(e^{2x}\cdot y\right)$$


Second, some mention of the product rule would make it clear to the teacher where you got the idea to go from line $1$ to line $2$.

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