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I had some homework for my differential equations class, and one of the questions completely stumped me, reproduced here:

Find $\frac{dy}{dx}$ for $y = ce^{-x} + e^{-x}\int_0^x\frac{tan(t)}{t}dt$

My next line looked like

$\frac{dy}{dx} = -ce^{-x} + \frac{d}{dx}(e^{-x}\int_0^x\frac{tan(t)}{t}dt)$

and by using the Fundamental Theorem of Calculus and the Product Rule (I don't know how correctly), my next line looked like

$\frac{dy}{dx} = -ce^{-x} + (e^{-x}\frac{d}{dx}\int_0^x\frac{tan(t)}{t}dt - e^{-x}\int_0^x\frac{tan(t)}{t}dt )$

and subsequently

$\frac{dy}{dx} = -ce^{-x} + e^{-x}\frac{tan(x)}{x} - e^{-x}\int_0^x\frac{tan(t)}{t}dt$

I tried evaluating the integral, since a simple derivation made it clear that it couldn't be avoided. However, I was not able to do so, and when looking it up on www.symbolab.com, it turns out to have no elementary antiderivative/is non-integrable. Does anyone know how to solve the original question?

NOTE: the assignment deadline already passed and I have already been graded on my attempt at this question. This is not an attempt to pass off anyone's insights and work as my own.

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    $\begingroup$ That looks fine to me, have you asked your instructor/teacher why you got it wrong? $\endgroup$
    – Andrew Li
    Commented Feb 28, 2018 at 0:17
  • $\begingroup$ @AndrewLi I only just received the assignment back. I haven't had the chance yet. Is it fine to leave the final answer in integral form? The second part of the question is if y(0) = 3, find c $\endgroup$
    – arnavlohe15
    Commented Feb 28, 2018 at 0:25
  • $\begingroup$ Why is the derivative of an integral function not allowed to also be an integral function in this question? $\endgroup$
    – AlkaKadri
    Commented Feb 28, 2018 at 0:25
  • $\begingroup$ @arnavlohe15 If there's no elementary integral, I'd expect it to be acceptable. You've applied the product rule and FTC correctly. $\endgroup$
    – Andrew Li
    Commented Feb 28, 2018 at 0:26
  • $\begingroup$ @arnavlohe15 Also, as you mention the second part of the question, it tells me that your instructor doesn't expect you to integrate because $\int_0^0 f(x)dx = 0$ so it seems even more valid to leave the integral in the derivative. $\endgroup$
    – Andrew Li
    Commented Feb 28, 2018 at 0:29

1 Answer 1

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Since $$y = c e^{-x} + e^{-x} \int_0^x \frac{\tan t}{t} \, dt,\tag1$$ as you correctly show $$y' = -ce^{-x} + e^{-x} \frac{\tan x}{x} - e^{-x} \int_0^x \frac{\tan t}{t} \, dt.\tag2$$

Now if we rearrange (1) we have $$e^{-x} \int_0^x \frac{\tan t}{t} \, dt = y - c e^{-x},$$ so substituting this result into (2) we have $$y' = -c e^{-x} + e^{-x} \frac{\tan x}{x} - (y - ce^{-x}),$$ or $$y' + y = \frac{e^{-x} \tan x}{x},$$ a first-order differential equation that is free from any integral sign and has (1) as a solution.

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