Use Frobenius' method to find two independent solutions to the ODE $4tx''(t)+2x'(t)+x(t)=0$

Question

It is known that $t=0$ is a regular singular point of

$4tx''(t)+2x'(t)+x(t)=0$.

By Frobenius' method, show that two independent solutions of the ODE are given by

$x_1(t)=\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}t^k$ and $x_2(t)=\sqrt{t}\sum_{k=0}^\infty$$\frac{(-1)^k}{(2k+1)!}t^k$

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So far I have used the Frobenius ansatz $x(t)=\sum_{k=0}^\infty a_kt^{k+r}$ which implies

$x'(t)=\sum_{k=0}^\infty (k+r)a_kt^{k+r-1}$ and

$x''(t)=\sum_{k=0}^\infty (k+r-1)(k+r)a_kt^{k+r-2}$

Substituting into the ODE gives

$4\sum_{k=0}^\infty (k+r-1)(k+r)a_kt^{k+r-1}+2\sum_{k=0}^\infty (k+r)a_kt^{k+r-1}+\sum_{k=0}^\infty a_kt^{k+r}=0$

We can then change the summation index to make all of the powers of $t$ in each summation equal to $k+r-1$:

$4\sum_{k=0}^\infty (k+r-1)(k+r)a_kt^{k+r-1}+2\sum_{k=0}^\infty (k+r)a_kt^{k+r-1}+\sum_{k=1}^\infty a_{k-1}t^{k+r-1}=0$

Which we can combine to make:

$4(r-\frac{1}{2})ra_0t^{r-1}+\sum_{k=1}^\infty t^{k+r-1}(a_k(4(k+r-\frac{1}{2})(k+r))+a_{k-1})$

It is here that I am unsure of what to do. I believe I need to find the indicial equation? Any help would be much appreciated!

Answer 1

$$\color {blue}{4(r-\frac{1}{2})ra_0t^{r-1}}+\sum_{k=1}^\infty t^{k+r-1}(a_k(4(k+r-\frac{1}{2})(k+r))+a_{k-1})=0$$

Yes you need to find the indicial equation. Take the coefficient of the lowest power of $t$. You should get for all $t$:

$$t^ {r-1} (4ra_0(r-1)+2ra_0)=0$$ $$2ra_0(2r-1)=0$$ $$r(2r-1)=0$$ $$\implies r=0,r=\frac 12$$

What you have done is correct. Now you have the values of $r$ you can continue.

Next step: you have to find the recurrence relation and deduce, if it is possible, a recurrence formula and find the series solution.

Answer 2

To find the indicial equation, you begin by putting it in standard form: $$x'' + (1/2t)x' + (1/4t)x = x'' + p(t)x' + q(t)x = 0$$

The indicial equation is $$r(r-1) + p_0r + q_0 = 0$$ where $$ p_0 = \lim_{t \to 0}t \,p(t) \\ q_0 = \lim_{t \to 0} t^2 \, q(t) $$