Basic Reference material about ODEs such as saparability with calculations and examples?

Question

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.

Answer 1

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.

Answer 2

Hint:

Let $r=x+1$ ,

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

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

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

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

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

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

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

$r\dfrac{d^2u}{dr^2}-\dfrac{du}{dr}=\dfrac{u^2}{2}+C_1$