Need help with a complex differential equation

In summary, the equation looks like it could be 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'(
  • #1
Kvad
18
0
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 $$n≥3,\ p<x<1$$
 
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  • #2
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?
 
  • #3
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...
 
  • #4
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?)
 
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  • #5
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.
 
  • #6
Kvad said:
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?
 
  • #7
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...
 
  • #8
In that case I cannot help you.
Good luck.
 

1. What is a complex differential equation?

A complex differential equation is a mathematical equation that involves derivatives of a complex-valued function. It is used to model systems that involve complex quantities, such as electrical circuits, fluid dynamics, and quantum mechanics.

2. Why is help needed with a complex differential equation?

A complex differential equation can be difficult to solve, especially if it is a higher-order equation or if it involves complex functions. Therefore, seeking help from a mathematician or a specialized software can be beneficial in finding a solution.

3. What are some common techniques for solving complex differential equations?

Some common techniques for solving complex differential equations include separation of variables, integrating factors, and power series solutions. Depending on the complexity of the equation, other methods such as Laplace transforms, numerical methods, and computer simulations may also be used.

4. Can complex differential equations be applied to real-world problems?

Yes, complex differential equations are widely used in various fields of science and engineering to model and understand real-world phenomena. They are particularly useful in systems that involve oscillations, waves, and dynamic behavior.

5. How can I improve my understanding of complex differential equations?

To improve your understanding of complex differential equations, it is important to have a solid foundation in calculus, complex analysis, and linear algebra. Practice solving different types of equations and seek guidance from experts or online resources when needed. It may also be helpful to study specific applications of complex differential equations in your field of interest.

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