Need help with a complex differential equation.

[Kvad]

Hi everyone!

Have been trying to solve this equation for a while, but just cannot find the right way. Could you, please, help on this one? Thanks!

\(\displaystyle \color{black}{\dfrac{x^{n-1}\, -\, y(x)^{n-1}}{n\, -\, 1}\, -\, \dfrac{x^{n-2}\, -\, y(x)^{n-2}}{n\, -\, 2}p\, =\, \dfrac{1}{2}y(x)^{n-3}\left(p\, -\, y(x)\right)^2\,y'(x)}\)

\(\displaystyle \color{black}{\mbox{Where }\, n\, \geq\, 3\, \mbox{ and }\, 0\, <\, p\, <\, x\, <\, 1,\, n,\, p\, -\, \mbox{ parameters.}}\)

 

[DrPhil]

Hi everyone!

Have been trying to solve this equation for a while, but just cannot find the right way. Could you, please, help on this one? Thanks!

I get confused easily - would rather just type \(\displaystyle y\) instead of \(\displaystyle y(x)\).

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

Where \(\displaystyle n \ge 3 \) and \(\displaystyle 0 < p < x < 1\)

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

Or maybe it would be better to write it as

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

and treat the left side = 0 as a homogeneous equation, and the right side particular?

What have you tried? I'll have to admit that I generally opt for numerical solutions, but that is awkward when there are parameters in the equation.

 

[Kvad]

Sorry, I actually had it with just "y", but decided to add x to be more comprehensive

I have tried various transformations and substitutions, but they did not help eventually in simplifying further the expression, something like z=y^(n-2) etc... I do not have much knowledge in differential equations, but usually in homogeneous equations we substitute y=Fx, can it help in this particular example?

The solution should be the same for any n or p. Say, for n=4 it with some rearrangements it becomes like this. Is it easier to see a way for solution here? (EDIT: the equation is equal to 0)

. . . . .\(\displaystyle \color{black}{3(p\, -\, y)^2\, y' y\, +\, x^2(2x\, -\, 3p)\, +\, 3py^2\, -\, 2y^3}\)