Need help with a complex differential equation.

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Kvad

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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!

xn−1 − y(x)n−1n − 1 − xn−2 − y(x)n−2n − 2p = 12y(x)n−3(p − y(x))2 y′(x)\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)}

\(\displaystyle \color{black}{\mbox{Where }\, n\, \geq\, 3\, \mbox{ and }\, 0\, <\, p\, <\, x\, <\, 1,\, n,\, p\, -\, \mbox{ parameters.}}\)
 
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Kvad said:
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!
Click to expand...
I get confused easily - would rather just type y\displaystyle y instead of y(x)\displaystyle y(x).

xn−1−yn−1n−1−xn−2−yn−2n−2p=12 yn−3 (p−y)2 y′\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 n≥3\displaystyle n \ge 3 and 0<p<x<1\displaystyle 0 < p < x < 1

12 yn−3 (p−y)2 y′+yn−1n−1−yn−2n−2p = xn−1n−1−xn−2n−2p\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

12 yn−3 (p−y)2 y′−xn−1n−1+xn−2n−2p = −yn−1n−1+yn−2n−2p\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.
 
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)

. . . . .3(p − y)2 y′y + x2(2x − 3p) + 3py2 − 2y3\displaystyle \color{black}{3(p\, -\, y)^2\, y' y\, +\, x^2(2x\, -\, 3p)\, +\, 3py^2\, -\, 2y^3}
 
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