After this substitution I got to the point

$$\cosh^6 (\theta)y'-\sinh(2\theta)-\cosh^4 (\theta)=0$$

and now let

$$z=\cosh^2 (\theta)$$

so $$z^3 y'-z^2-\sinh(2\theta)=0$$

but then I started to question my substitution, look at the graph here. Is the substitution even possible? Can we actually even get $\textrm{arcsinh}(x)=\textrm{arccosh}(y)$? From the graph in WA, it seems like the hypergeometric sin/cosh do not even cross so the substition in the title is wrong or?

Again from the book, now on page 644.

  • $\begingroup$ Which book are you talking about? $\endgroup$
    – Pedro
    Feb 17, 2012 at 17:07
  • $\begingroup$ @Peter: the foreign course book here, not available in English (sorry). $\endgroup$
    – hhh
    Feb 17, 2012 at 17:11
  • $\begingroup$ I gave you a solution but I don't know how your solution would work. I'll think about it and edit ASAP. $\endgroup$
    – Pedro
    Feb 17, 2012 at 17:21
  • $\begingroup$ Solution from my uni is here, p. 636-637 (course book) mention something called "variation of constant" -- I have never understood the term, apparently just first-order linear DY? For future random walkers, I instruct to read the answers and the wikipedia -- I think the book is incomprehensible at that point. $\endgroup$
    – hhh
    Feb 19, 2012 at 10:15

2 Answers 2


I'll try to explain the idea of the "varioidaan vakiota" that is stated in your .pdf. The idea traces back to Legendre, which used it to solve linear ODEs.

We have the equation


We first solve the homogeneous equation




$$\log y = \log(1+x^2)+C_1$$

$$ y = C(1+x^2)$$

What Legendre thinks now, which Spiegel says "in first sight, seems ridiculous is":

Let's assume $C$ is not constant, but rather variable. What will this new function $C(x)$ be so that

$$y = C(x)(1+x^2)$$

is a solution to our original equation?

So in your case we have that, differentiating produces

$$ y' = C'(x)(1+x^2)+C(x) 2x$$

But from our equation we have that $y'=\dfrac{2xy}{1+x^2}+(1+x^2)$ and that $ \dfrac{y}{1+x^2} = C(x)$

So plugging this in we have

$$y' = C'(x)(1+x^2)+\frac{2xy}{1+x^2}$$

So that, comparing to our original equation we have $C'(x) = 1$ or $C(x) = x+C_1$, so that our solution is

$$y = (x+C_1)(1+x^2)$$

Hope this helped clear out the "varioidaan vakiota" issue. For more infor refer to Spiegel's book on Differential Equations, page 202.

You have

$$(1+x^2)y' -2xy = (1+x^2)^2$$


$$(1+x^2)\dfrac{dy} {dx} -2xy = (1+x^2)^2$$

$$\dfrac{dy} {dx} -\dfrac{2x}{1+x^2}y = 1+x^2$$

You can solve this by the integrating factor $$\exp\left(-\log\left(1+x^2\right)\right)=\dfrac{1}{1+x^2}$$ which will give

$$\dfrac{1}{1+x^2}\dfrac{dy} {dx} -\dfrac{1}{1+x^2}\dfrac{2x}{1+x^2}y = 1$$

$$\left(\dfrac{y}{1+x^2}\right)' = 1$$

$$\dfrac{y}{1+x^2} = x+C$$ $$y = (x+C)(1+x^2)$$

  • $\begingroup$ ...sorry what does the last sentence actually mean? Can you solve it that way? $\endgroup$
    – hhh
    Feb 19, 2012 at 0:17
  • $\begingroup$ @hhh I've been looking at the problem and the substitution is IMO a waste of time, since what it only helps is into seeing what integrating factor to use, which is $$\frac{1}{{{{\cosh }^2}\theta }} = \frac{1}{{1 + {x^2}}}$$ so the problem is solved in the same manner either way. If you want I can order my thoghts on the substitution and post it. $\endgroup$
    – Pedro
    Feb 19, 2012 at 0:33
  • $\begingroup$ The "integrating factor" is my weakest link for sure, not sure how with different differentials. I think it would be good idea to somehow clarify the final sentence, now it is a bit separate from the other part of the answer. $\endgroup$
    – hhh
    Feb 19, 2012 at 0:46
  • $\begingroup$ @hhh Added info on your doubt. Any problem, let me know. $\endgroup$
    – Pedro
    Feb 20, 2012 at 0:10

Hint: this is a linear differential equation. In your case, it may be wise to skip the beginning of the Wikipedia article and go directly to "first order equation".

  • $\begingroup$ ...any idea why my school solution mentions the method "variation of parameters" needed to solve the problem? I think they referring to this here by the term "variation of constant" (free translation). I can get the same solution without at least explicitly using some variation of parameters -method. $\endgroup$
    – hhh
    Feb 19, 2012 at 10:22
  • $\begingroup$ Moved this point to this question here to keep questions specific. $\endgroup$
    – hhh
    Feb 19, 2012 at 10:29

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