Need help with a complex differential equation

[Kvad]

Hi guys!

I am really stuck at quite a complicated (from my point of view) differential equation. I would really appreciate any hints or suggestions on how to tackle and solve it if it is possible, thanks!


$$(\frac{x^{n-1}-y^{n-1}}{n-1})-(\frac{x^{n-2}-y^{n-2}}{n-2})p=\frac{1}{2}y^{n-3}(p-y)^2y'(x)$$

p and n - parameters

where $$n≥3,\ p<x<1$$

 

[Simon Bridge]

Have you tried:
1. putting the LHS over a common denominator?
2. putting n=3,4,5 ... writing it out and seeing if you recognize the form of the equation?

 

[Kvad]

Yes, and the equation is more or less solvable when n=3, but starting from n=4 there is no more difference of squares which allowed for simplification... Say, for n=5 we have:
$$\frac{x^3(3x-4p)+4py^3-3y^4}{y'(x)}-6y^2(p-y)^2=0$$

or the general form of the same equation:

$$\frac{x^{n-2}((n-2)x-(n-1)p)+(n-1)py^{n-2}-(n-2)y^{n-1}}{y'(x)}-\frac{(n-2)(n-1)}{2}y^{n-3}(p-y)^2=0$$

But it still does not shed any light for me on possible ways to find a solution...

 

[Simon Bridge]

Usually you'd have to explore a bit more before light dawns.
What level are you doing this at?

I am a bit puzzled as to why you keep putting the y' in the denominator.

The general form seems to be something like f(x)+g(y)=h(y)y' right?
Or s(y)y' = p(x)/q(y)+1 ... both nearly things you know strategies for perhaps?

What happens if you change variables:

$z=y^{n-2}$ then $z'=(n-2)y^{n-3}y'$

... looks close to something useful.

Also
- why is it important that p<x<1 ? (may be a clue - can you exploit this restriction?)
- is n an integer?
- is p > 0 too? (i.e. can x be negative?)

 

[Kvad]

I am sorry but I did not understand the question about "what level am I doing it at", could you, please, specify?

As for the general form you are right, f(x)+g(y)=h(y)y' is the one! I do not know much about differential equations, could you, please, refer me to a method, which deals with such forms of equations?

Again, changing variables seems promising, but the squared term (p-y) distorts the picture...

For the questions:
-This restriction applies from the set up this equation was derived, I do not think it can be a clue in a mathematical solution, but I cannot completely reject such possibility...
- Yes, n is an integer.
- Yes, p is strictly positive and less than x, y is also strictly positive.

 

[Simon Bridge]

I am sorry but I did not understand the question about "what level am I doing it at", could you, please, specify?

Education level.

Is this for a course at some sort of school or university or college or something?

As for the general form you are right, f(x)+g(y)=h(y)y' is the one! I do not know much about differential equations, could you, please, refer me to a method, which deals with such forms of equations?

There is no one method, just a bunch of strategies.
IF it were y'=f(x)/g(y) you'd have no problem though right?

Again, changing variables seems promising, but the squared term (p-y) distorts the picture...

Yes - you'll probably have to look for another kind of substitution.

Can you find one where z'=h(y)y' ? Would that help?
Do you see the kind of thinking that is needed now?

For the questions:
-This restriction applies from the set up this equation was derived, I do not think it can be a clue in a mathematical solution, but I cannot completely reject such possibility...
- Yes, n is an integer.
- Yes, p is strictly positive and less than x, y is also strictly positive.

It's always a clue.

What is the setup?

 

[Kvad]

Thank you so much, Simon, for all your suggestions and hints! I really appreciate it, but unfortunately I cannot say more about the problem beyond its mathematical formulation...

 

[Simon Bridge]

In that case I cannot help you.
Good luck.