# First order linear ODE solution

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

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.

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