Given the following ODE

$$ 2\left[1 - x - b\right]y' -2a(1-x)y + x^{a+1}=0 $$

I am trying to find a solution for $y(x)$. My approach has been to firstly rearrange the equation in the canonical form

$$ y' - \frac{a(1-x)}{1-x-b}y = - \frac{z^{a+1}}{2(1-x-b)} $$

Then, multiply by the following integrating factor

$$ \mu(x) = \exp \left(-\int \frac{a(1-x)}{1-x-b}dx\right) = (1-x-b)^{ab}e^{-ax} $$

This allows me to write the equation in the form

$$ \frac d{dx} \left((1-x-b)^{ab}e^{-ax} y\right) = - \frac{x^{a+1}}{2(1-x-b)} (1-x-b)^{ab}e^{-ax} $$

It is at this point that I am not 100% sure on how to proceed. Integrating both sides, I have an integral that looks like a Gamma integral

$$ \Gamma(z) = \int^\infty_0 t^{x-1}e^{-t}dt $$

I have the boundary condition that $y(1)=1$, so I think I can fix the integration limits between $\int^1_x$, however, I'm struggling to make the final step.


1 Answer 1


Every first order differential equation in the form of $ay'+by+c=0$ (where $a , b, c, y$ are functions of $x$) can be written as $\frac{d}{dx}\left({\mu y}\right)=-c\mu$, so in your case,
$$\mu=(1-x-b)^{ab}e^{-ax}, b=\frac{-a(1-x)}{1-x-b}, c=\frac{x^{a+1}}{2(1-x-b)}$$

If we rearrange the statement $\frac{d}{dx}\left({\mu y}\right)=-c\mu$, we get
$$y=\frac{-\int c\mu\space dx}{\mu}$$

But, when entered in Wolfram Alpha , it says that the integral $\int c\mu\space dx$ has no solution in terms of standard mathematical functions, hence y cannot be expressed in terms of standard mathematical functions.

Therefore, no solution of y exists.

  • $\begingroup$ @TShiong how did you do such a level of formatting? How did you center the y = - integral stuff...? $\endgroup$
    – Zayaan
    Jul 29, 2023 at 17:32
  • $\begingroup$ You need learn formatting in LaTex. $\endgroup$
    – TShiong
    Jul 29, 2023 at 21:08
  • $\begingroup$ But I did learn, but still dont know how to do that. Can you recommend a good website? $\endgroup$
    – Zayaan
    Jul 30, 2023 at 9:55

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