I am trying to show this kind of non-linear $y''''=y'y''/(1+x)$ in normal form. For example here if $y=e^{x}\rightarrow y^{(n)}=e^{x}\rightarrow x=-1$, where $y^{(n)}$ means $n$th differential, then $x=-1$, too weak idea. When I google with differential or anything like that, most of the material does not look the material that I need. I need to solve different type of problems such as this homework

$$\begin{cases} u'=(u+v)^{2} \\ v''=x+u'v' \end{cases} $$

I am not requesting you to solve them but I am requesting some material because I find my book quite hard-reading in this section. The earlier chapter begun that something is something, without much further ado really why?, and now the next advanced chapters are referring to the past chapters. The idea is a rush introduction to this topic in an engineering course so I think it explains quite a bit about the pedagogy.

  • $\begingroup$ Which book are you using now? Also, the usual notation is $y^{(n)}$ for the $n^\text{th}$ derivative. For the first problem, you might need a series solution. $\endgroup$
    – bgins
    Feb 14, 2012 at 9:10
  • $\begingroup$ @bgins: reading the same old foreign book as in my earlier questions about Indefinite integral, now in the section about differentials, sorry I wish I had the book as online or English version... I does cover some theory but not introduce the methods to really solve this kind of things, very well it is thick (1k pages and quite inaccessible)... $\endgroup$
    – hhh
    Feb 14, 2012 at 9:19

2 Answers 2


You might want to check out David Joyner's excellent 2007 (draft) book.

One method that would work for the first equation, $y^{(4)}=\frac{y'y''}{1+x}$, would be the power series or series solution or Frobenius method. The method consists of assuming a Taylor/Maclaurin series form: $$ y=\sum_{n=0}^\infty a_nx^n $$ which (after adjusting the summation variable $n$ to ignore the leading terms which are zero) has $k^\text{th}$ derivative $$ y^{(k)}=\sum_{n=0}^\infty\frac{(n+k)!}{n!}a_{n+k}x^n $$ so that, equating terms with the same power of $x$, the differential equation becomes a recurrence relation on the sequence $\{a_n\}$ of coefficients, with the first few ($4$ in our case) free (to match initial condidions) and the rest determined: $$ \sum\frac{(n+4)!}{n!}a_{n+4}x^n= \left(\sum\frac{(n+2)!}{n!}a_{n+2}x^n\right) \left(\sum(n+1)a_{n+1}x^n\right) \left(\sum(-1)^nx^n\right) $$ $$ \forall{n}:\quad a_{n+4}=\frac{n!}{(n+4)!} \sum_{j=0}^n(-1)^{n-j} \sum_{k=0}^j (k+2)(k+1) (j-k+1) a_{k+2} a_{j-k+1} $$ where the summations above with no indices specified are for $n\ge 0$ and the coefficients of the product of two series were found with a kind of convolution formula, $$ \left(\sum_{n\ge0}a_nx^n\right) \left(\sum_{n\ge0}b_nx^n\right)= \sum_{n\ge0}\left(\sum_{k=0}^n a_kb_{n-k}\right)x^n. $$ So for instance $$a_4=\frac{2a_1a_2}{4!}$$ $$a_5=\frac{2a_2(3a_3+2a_2-a_1)}{5!}$$ and so on. You need to be able to estimate the growth of the $a_n$ and use a ratio or root test to determine the radius of convergence of your series, and in some cases, you will be able to find an analytic solution.

  • $\begingroup$ ...is this term $\left(\sum(-1)^nx^n\right)$ for the divisor $1+x$? $\endgroup$
    – hhh
    Feb 14, 2012 at 23:02
  • $\begingroup$ yes, exactly... $\endgroup$
    – bgins
    Feb 14, 2012 at 23:48
  • $\begingroup$ When I read your answer, I think my book (p.427) has a mistake: it apparently claims that "the exponent function of the form" $E(X)=\sum_{k=0}^{\infty}\frac{(ax)^{k}}{k!}$ "is an unique solution to any initial-valued DY problem" and then it in a rush of sentence states that $e^{x}$ some sort of basic solution. Sorry the book actually states nothing, it more like outlines things and then requires one to calcutale things...feeling that a lot of gaps here. $\endgroup$
    – hhh
    Feb 15, 2012 at 0:11
  • $\begingroup$ How do you deal with divisors if you have them such as $1+x+y'''$ or $1+x+x^{2}+y'''+y^{(5)}$? Like the first one, I cannot immeadiately see a geometric serie -- perhaps to separate it if with the above case and then just one-by-one use the convolution rule for every part? $\endgroup$
    – hhh
    Feb 15, 2012 at 0:32


Let $r=x+1$ ,

Then $\dfrac{d^4y}{dr^4}=\dfrac{1}{r}\dfrac{dy}{dr}\dfrac{d^2y}{dr^2}$


Let $u=\dfrac{dy}{dr}$ ,

Then $r\dfrac{d^3u}{dr^3}=u\dfrac{du}{dr}$

$\int r\dfrac{d^3u}{dr^3}~dr=\int u\dfrac{du}{dr}~dr$

$\int r~d\left(\dfrac{d^2u}{dr^2}\right)=\int u~du$

$r\dfrac{d^2u}{dr^2}-\int \dfrac{d^2y}{dr^2}~dr=\int u~du$



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