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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.

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    Solution from my uni is [here](https://noppa.aalto.fi/noppa/kurssi/mat-1.1020/viikkoharjoitukset/Mat-1_1020_mallit_av7.pdf), 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.2012-02-19

2 Answers 2

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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

y'-\frac{2xy}{1+x^2}=1+x^2

We first solve the homogeneous equation

y'-\frac{2xy}{1+x^2}=0

y'=\frac{2xy}{1+x^2}

\frac{y'}{y}=\frac{2x}{1+x^2}

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

$ 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

or

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

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

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

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

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    @hhh Added info on your doubt. Any problem, let me know.2012-02-20
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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".

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    Moved this point to this question [here](http://math.stackexchange.com/questions/110884/variation-of-constant-method-to-solve-linear-dys) to keep questions specific.2012-02-19