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I am wondering how to find an explicit, closed-form solution for the following first-order differential equation:

$${y}' = \cosh^{-1}\left ( x \right ) + \mathrm {Si}(x)$$

Where $\mathrm {Si}(x)$ denotes the sine integral, and is defined as the following: $$\mathrm {Si}(x) = \int_0^x \frac {\sin {t}} {t} \mathrm{d}t$$

My initial course of action would be to find an integrating factor.

First, I would rewrite the above equation into the following form:

$${y}' + p(x) = q(x)$$

I'll set the following:

$$p(x) = -\mathrm {Si}(x)$$ $$q(x) = \cosh^{-1}(x)$$

Therefore:

$${y}' - \mathrm {Si}(x) = \cosh^{-1}(x)$$

I then set: $$ \mu(x) = e^{-\int \mathrm{Si}(x) \mathrm{d}x} = e^{x \mathrm{Si}(x) + \cos x}$$

Multiplying through the differential equation by the integrating factor I obtain:

$$ e^{x \mathrm{Si}(x) + \cos x } {y}' - e^{x \mathrm{Si}(x) + \cos x } \mathrm{Si}(x) = e^{x \mathrm{Si}(x) + \cos x } \cosh^{-1}(x) $$

I can now write the following:

$$ y(x) = \frac {1} {e^{x \mathrm{Si}(x) + \cos x}} \int_a^x e^{s \mathrm{Si}(s) + \cos s} \cosh^{-1}(s) \mathrm{d}s$$

This is where I get stuck: I have no idea how to solve that integral. I'm not even sure if this was the right approach to take for this problem as I'm relatively new to differential equations. Any advice would be appreciated. I am particularly interested in developing a deeper understanding of the methods behind solving this as opposed to simply learning a procedure.

Wolfam Alpha claims the following is a closed-form solution for this differential equation:

$$ y(x) = c_1 + x\mathrm{Si}(x)-\sqrt{x-1}\sqrt{x+1}+\cos x + x \cosh^{-1}(x) $$

I recognize the presence of the antiderivative of $\mathrm{Si}(x)$ in this solution, but I'm mostly at a loss as to where the rest came from.

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    $\begingroup$ Why can't you just integrate both sides? $\endgroup$ – Amzoti Nov 14 '14 at 3:22
  • $\begingroup$ Wow, I don't know where my brain is today. I made this more complicated than it needed to be. Thanks! $\endgroup$ – SCS Nov 14 '14 at 3:24
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    $\begingroup$ That happens to me at least ten times a day! :-) By the way, see item $8$ here for the integration of $Si(x)$: mathworld.wolfram.com/SineIntegral.html $\endgroup$ – Amzoti Nov 14 '14 at 3:25
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Thank you to Amzoti for his comment -- it turns out all you have to do is integrate both sides. It makes sense, in retrospect. The radical terms looked vaguely familiar, and now I remember that they are part of the antiderivative for $ \cosh^{-1}(x) $.

$$ {y}' = \cosh^{-1}(x) + \mathrm{Si}(x) $$

$$ \int {y}' \mathrm{d}y = \int \left[\cosh^{-1}(x) + \mathrm{Si}(x)\right] \mathrm{d}x $$

$$ y = x \cosh^{-1}(x) - \sqrt{x-1}\sqrt{x+1} + x \mathrm{Si}(x) + \cos x + C $$

Which matches the solution from Wolfram Alpha I provided in my question.

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