I am trying to show this kind of non-linear y''''=y'y''/(1+x) in normal form. For example here

Mary Buchanan

Mary Buchanan

Answered question

2022-01-22

I am trying to show this kind of non-linear y=yy1+x in normal form. For example here if y=exy(n)=exx=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
{u=(u+v)2v=x+uv
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 & Explanation

Papilys3q

Papilys3q

Beginner2022-01-22Added 34 answers

One method that would work for the first equation, y(4)=yy1+x, would be the power series or series solution or Frobenius method. The method consists of assuming a Taylor/Maclaurin series form:
y=n=0anxn
which (after adjusting the summation variable n to ignore the leading terms which are zero) has kth derivative
y(k)=n=0(n+k)!n!an+kxn
so that, equating terms with the same power of x, the differential equation becomes a recurrence relation on the sequence an of coefficients, with the first few (4 in our case) free (to match initial condidions) and the rest determined:
(n+4)!n!an+4xn
=((n+2)!n!an+2xn)((n+1)an+1xn)((1)nxn)
n:an+4=n!(n+4)!j=0n(1)nj
k=0j(k+2)(k+1)(jk+1)ak+2ajk+1
where the summations above with no indices specified are for n0 and the coefficients of the product of two series were found with a kind of convolution formula,
(n0anxn)(n0bnxn)=n0(k=0nakbnk)xn.
So for instance a4=2a1a24!
a5=2a2(3a3+2a2a1)5!
and so on. You need to be able to estimate the growth of the an 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.
limacarp4

limacarp4

Beginner2022-01-23Added 39 answers

Hint:
Let r=x+1,
Then d4ydr4=1rdydrd2ydr2
rd4ydr4=dydrd2ydr2
Let u=dydr,
Then rd3udr3=udydr
rd3udr3dr=ududrdr
rd(d2udr2)=udu
rd2udr2d2ydr2dt=udu
rd2udr2dudr=u22+C1

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