3
$\begingroup$

Edit: from the answers, I have learnt that the differential equation can be solved by expressing it as being a hypergeometric differential equation. My question now is that, how many a function in the form of
$x(1-x)^2y''(x)+(1-x)^2y'(x)+ay(x)=0$
be transformed into the form of
$\eta(1-\eta)f''(\eta)+(b-c\eta)f'(\eta)+df(\eta)=0$?

Original question: How may one solve a differential equation in the form of:
$\frac{dy}{dx}=P(x) -ky^2$
I have attempted at reducing it into a second order homogeneous equation in the form of $\frac{d^2u}{dx^2}=-kP(t)u$
by making the substitution of
$ky= \frac{\frac{du}{dx}}{u}$
However, I am still unable to solve this.
Are there any methods for solving either equation?

If it helps, P(x) is the derivative of:
$f(x)=\frac{a-be^{cx+d}}{1-e^{cx+d}}$ where a,b,c,d are constants
Additionally, y=0 when x=0, and y=0 as x$\rightarrow$ infinity

A numerical approach to solving the equation with randomly chosen values for constants substituted in gives the following graph: Link
which looks like (maybe) a chi-square distribution....?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ is $$P(x)=-\frac{c (b-a) e^{c x+d}}{\left(e^{c x+d}-1\right)^2}$$? $\endgroup$ – Dr. Sonnhard Graubner Jun 23 '17 at 10:14
  • $\begingroup$ yes indeed it is $\endgroup$ – Samus Jun 23 '17 at 15:00
2
$\begingroup$

The equation you are working with is the Ricatti equation: $$y'+ky^{2}=P(x)$$ Where $$P(x)=\frac{d}{dx}\frac{a-be^{cx+d}}{1-e^{cx+d}}=\frac{c(a-b)e^{cx+d}}{(1-e^{cx+d})^{2}}$$ By doing your substitution $$y(x)=\frac{1}{k}\frac{\frac{du(x)}{dx}}{u(x)}$$ you indeed have the second order linear ode $$\frac{d^{2}u}{dx^{2}}=kP(x)u(x)=k\frac{c(a-b)e^{cx+d}}{(1-e^{cx+d})^{2}}u(x)$$ First it is nice to clean the equation a little bit by letting $z=c{x}+d$ and $k(a-b)/c=\alpha$, then $$\frac{d^{2}u(z)}{dz^{2}}=\alpha\frac{e^{z}}{(1-e^{z})^{2}}u(z)$$ Then you may also want a change of variables $$\xi=e^{z}$$ So that $$\xi(1-\xi)^{2}\frac{d^{2}u(\xi)}{d\xi^{2}}+(1-\xi)^{2}\frac{du(\xi)}{d\xi}-\alpha{u}(\xi)=0$$ Then you let $1-\sqrt{4\alpha+1}=\gamma$ and do the following substitutions $$\sigma=(x-1)$$ $$u(\xi)=\sigma^{\frac{1-\gamma}{2}}f(\sigma)$$ To give the Gauss hypergeometric ode $$\sigma(1-\sigma)\frac{d^{2}}{d\sigma^{2}}f(\sigma)+(\gamma+(1-\gamma)\sigma)\frac{d}{d\sigma}f(\sigma)-\frac{1}{4}\gamma^{2}f(\sigma)=0$$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ why is it that $(1-e^{cx+d})^2$ becomes $1-e^z$? $\endgroup$ – Samus Jun 23 '17 at 15:17
  • $\begingroup$ Yes, that is the mistake, I'll edit that one $\endgroup$ – Kiryl Pesotski Jun 24 '17 at 9:06
  • $\begingroup$ Sorry, but what is $\sigma$ in the substitution? I believe there might be a typo as (x-1) does not really work there. Once again thank you so much for the help $\endgroup$ – Samus Jun 24 '17 at 14:31
  • 2
    $\begingroup$ Riccati was an Italian mathematician; it's important to spell his name correctly, because ricatti in Italian means blackmail. $\endgroup$ – egreg Jul 21 '17 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.