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I'm trying to solve the following differential equation by using the method of Frobenius. I'm however, having some trouble in doing so, I was hoping someone could help me out.

$2ty''+(1+t)y'-2y=0$

ATTEMPT: First of all, we need to control if we can actually use Frobenius' method. By looking at $2ty''$ we can easily conclude that $t=0$ is the only singular point of the differential equation. Then we need to control whether $t \cdot (1+t)$ and $ t^2 \cdot -2 $ are both analytic at $t=0$, which is true. Therefore, we can indeed use the method of Frobenius.

We let $y$ be of the form $y=\sum_{n=0}^{\infty}a_n \cdot t^{n+r}$, then we can get the following by differentiating:

$y=\sum_{n=0}^{\infty}a_n \cdot t^{n+r}$

$y'=\sum_{n=0}^{\infty}(n+r)a_n \cdot t^{n+r-1}$

$y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_n \cdot t^{n+r-2}$

We now plug these back in our differential equation:

$\begin{align} 2\sum_{n=0}^{\infty}(n+r)(n+r-1)a_n \cdot t^{n+r-1}+\sum_{n=0}^{\infty}(n+r)a_n \cdot t^{n+r-1} + \sum_{n=0}^{\infty}(n+r)a_n \cdot t^{n+r} - 2\sum_{n=0}^{\infty}a_n \cdot t^{n+r} =\\ \sum_{n=0}^\infty a_n[2(n+r)(n+r-1)+n+r]t^{n+r-1}+\sum_{n=1}^\infty a_{n-1}[n+r-3]t^{n+r-1} =\\ [a_0(2r(r-1))+ra_0]t^r+[2a_1(r+1)(r)+a_1(r+1)+a_0(r-2)]t^{r+1}+\sum_{n=2}^\infty[a_n[2(n+r)(n+r-1)+n+r] + a_{n-1}(n+r-3)]t^{n+r-1} = 0 \end{align}$ So then, out of the first term of the sum (since it has to be 0, as the sum is equal to zero), we can solve for r, which gives us $r=0$ or $r=\frac{1}{2}$, since $a_0$ is assumed to be non-zero. Otherwise it would just change the index of r.

Then, we set $a_0$ to be $1$. and we can solve for $a_0$ by using the second term. $r=0$ gives us that $a_1$ has to be $2$. $r= \frac{1}{2}$ gives us that $a_1$ has to be equal to $\frac{1}{2}$. The last term of the equation above gives us the recurrent relationship. I believe that when we have the recurrent relationship, by plugging in the two different $a_1$ values, we will get two linearly independent solutions for the differential equation and therefore we'll get the general solution of the differential equation. I'm not able however to find this recurrent relationship.

QUESTIONS: I'm having quite a bit of trouble with using this method to solve differential equations, so here are a couple of questions I want to ask:

  1. Is it always necessary to write out the first and second term of the sum, as to get the values for $r$ and $a_1$?
  2. Is it okay to just set $a_0$ equal to 1?
  3. In the sum, I chose to change the powers of t equal to n+r-1 as to factor these powers out, I have also set these to n+r in a previous attempt, however, that I couldn't solve. Is there a general approach as to how to set these powers? Do you always set them at the lowest power of t in the total differential equation?
  4. Is this a correct and efficient manner to solve such a problem with Frebenius' method? Or is there some work that could be omitted?

Thanks for your time,

K. Kamal

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We have: $$\sum_{n=0}^{\infty}a_n[2(n+r)(n+r-1)+n+r]t^{n+r-1} +\sum_{n=0}^{\infty}a_n[n+r-2]t^{n+r}=0$$ $$a_0[2r^2-2r+r]t^{r-1} + \sum_{n=1}^{\infty}[2(n+r)(n+r-1)+n+r]a_nt^{n+r-1}+\sum_{n=0}^{\infty}[n+r-2]a_nt^{n+r}=0$$ $$a_0[2r^2-r]t^{r-1}+\sum_{n=0}^{\infty}[2(n+r+1)(n+r)+n+r]a_{n+1}t^{n+r}+\sum_{n=0}^{\infty}[n+r-2]a_nt^{n+r}=0$$

Equating the coefficient of $t^{r-1}$, the lowest power of $t$ to be zero, we obtain $r =0, \frac12$. Equating the coefficient of $t^{n+r}$ to be zero, we have: $$(2(n+r)(n+r+1)+n+r)a_{n+1}+(n+r-2)a_n=0$$ $$\implies a_{n+1}=\frac{-(n+r-2)a_n}{2(n+r)(n+r+1)+n+r}\,\,, n =0, 1, 2,…$$

Hope you can take it from here by considering the two different values of $r$.

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  • $\begingroup$ You are saying to equate the coefficients of $t^{n+r}$ to zero, but in the equation you give, the second and the third term dont have the same power for $t$. Can you still equate the coefficients? $\endgroup$ – K.Kamal Dec 18 '17 at 12:49
  • $\begingroup$ @K.Kamal Sorry, My bad... $\endgroup$ – Rohan Dec 18 '17 at 13:03

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