1

Is there a method (or book) that treats differential equations of the following form: $$y'(x)+\alpha y(ax)+\beta y(x)=f(x),\alpha, a\in\mathbb{R}^*, \beta\in\mathbb{R}.$$ Thank you in advance.

Theory Nombre
  • 21
  • Welcome to Math StackExchange! First, differential equations are really hard to solve generally. Second, there might often not be any $y(x)$ that satisfies differential equations of the form you describe (so maybe no method has been developed). With this in mind, it's probably best if you give a specific example you're interested in for $\alpha$, $\beta$, and $f(x)$. – Sherwin Lott Aug 09 '20 at 17:50
  • My example is $y'(x)=4y(x/2)-y(x)+a$ with $a\in\mathbb{R}.$. – Theory Nombre Aug 09 '20 at 21:45
  • Letting $y(x)=x^{2}$ has no effect on the right-hand side but adds $2x$ to the left. Whereas, $y(x)=x$ adds $x$ to the right-hand side and $1$ to the left. Using these building blocks, I get the solution: $$y(x) = \frac{a}{2}x^{2} + ax$$ Perhaps this approach can be generalized. It would be nice if someone knew the name of this kind of differential equation (if there is one). – Sherwin Lott Aug 10 '20 at 15:07
  • Basically, we can add polynomials to the left-hand side of your equation: $$y'(x)+\alpha y(ax)+\beta y(x)=f(x),\alpha, a\in\mathbb{R}^*, \beta\in\mathbb{R}.$$ The polynomials being factors of: $1 + (\alpha a + \beta)x$, $2x + (\alpha a^{2} + \beta)x^{2}$, etc. Equating coefficients with $f(x)$, we can try and derive a taylor series for $y(x)$. – Sherwin Lott Aug 10 '20 at 16:33
  • Thank you. Yes, $y(x)=\frac{a}{2}x^2+ax$ is a solution. Are there other solutions? And, how did you manage to choose $y(x)=x^2$ directly? – Theory Nombre Aug 11 '20 at 21:30

1 Answers1

1

Let $y(x) = \sum\limits_{n=0}^{\infty}p_{n}x^{n}$ and $f(x) = \sum\limits_{n=0}^{\infty} q_{n}x^{n}$.

Then, the differential equation becomes:

$$\begin{align} y'(x)+\alpha y(ax)+\beta y(x) &= \sum\limits_{n=0}^{\infty}(n+1)p_{n+1}x^{n} + \sum\limits_{n=0}^{\infty}(a^{n} \alpha + \beta) p_{n}x^{n} \\ &= \sum\limits_{n=0}^{\infty}\left[(a^{n} \alpha + \beta) p_{n} + (n+1)p_{n+1} \right]x^{n} \\ &= \sum\limits_{n=0}^{\infty} q_{n}x^{n} \end{align}$$

This gives a relationship between the Taylor coefficients of $y(x)$ and $f(x)$ which we can equate to solve for the coefficients of $y(x)$.

If $p_{0} = y(0)$ is given, then we can uniquely solve for all the coefficients of $y(x)$ via the recursion:

$$p_{n+1} = \frac{q_{n} - (a^{n} \alpha + \beta) p_{n}}{n+1}$$

Sherwin Lott
  • 2,778