How to solve \(\displaystyle{\left({y}'\right)}^{{4}}{x}-{2}{y}{\left({y}'\right)}^{{3}}+{12}{x}^{{3}}={0}\)

abitinomaq1

abitinomaq1

Answered question

2022-04-01

How to solve
(y)4x2y(y)3+12x3=0

Answer & Explanation

Harry Gibson

Harry Gibson

Beginner2022-04-02Added 13 answers

Step 1
Set y(x)=12u(x2), then y(x)=xu(x2) and
x5[u(x2)]4x3u(x2)[u(x2)]3+12x3=0.
Cancelling x3, dividing by y(x)3, substituting x2=s and reordering gives the form
u(s)=su(s)+12u(s)3.
This now is a Clairaut DE with linear solution family
u(s)=Cs+12C3, y(x)=C2x2+6C3
You get one of the Maple solutions with C=1C1
The singular solution or envelope of u(s)=su(s)+f(u(s)) is obtained from 0=s+f(u(s)), here
0=s36[u(s)]4
The real solutions are
u(s)=±(36s)14
so that
u(s)=u(s)(s+12[u(s)]4)=±43s(36s)14
y(x)=±23x6|x|
This should work for x>0 as well as for x<0.
As usual, one can switch branches where solutions from the regular family touch the singular solution.
Theodore Davila

Theodore Davila

Beginner2022-04-03Added 14 answers

Let's look for polynomial solutions. If y=cxd+, then (y)4x2y(y)3=c4(d42d3)x4d3+. This x4d3 term can only cancel if d=2. So now try y=c2x2+c1x+c0 and it's not hard to see we must have c1=0 and c0c23=34

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?