Ordinary DEQ - Frobenius Method

Question

I have a DEQ:

$$y''-\frac{6}{x}y'+\frac{12+x^2}{x^2}y=0$$

Then in series form:

$$\sum_{m=0}^{\infty}(m+r)(m+r-1)a_mx^{m+r-2}-6\sum_{m=0}^{\infty}(m+r)a_mx^{m+r-2}+12\sum_{m=0}^{\infty}a_mx^{m+r-2}+\sum_{m=0}^{\infty}a_mx^{m+r}$$

The lowerst power is obviously $m+r-2$, and we get the indicial equation $r(r-1)-6r=0 \rightarrow r_1=0, r_1=7$

Next, we equate the powers, since the first 3 are the same, we change the index on the 4th series:

$$s_4=\sum_{m=2}^{\infty}a_{m-2}x^{m+r-2}$$

Rearranging the series:

$$((m+r)(m+r-1)-6(m+r)+12)a_m+a_{m-2}=0$$

Let $r=0$, i get:

$$a_m=-\frac{a_{m-2}}{m(m-7)+12}$$

Then $$a_2=-\frac{a_0}{30}, a_4=-\frac{a_2}{56}=-\frac{a_0}{1680}.....$$

now that just doesn't seem very likely,,i triple-checked my solution, and don't know where i might have made a mistake, can someone have a look?

Answer 1

The lowest power terms include the third term of the equation, you get $$ 0=r(r-1)-6r+12=(r-3)(r-4) $$ as indicial equation, which is then also compatible with the later recursion formula.

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The general solution now has the form $y(x)=x^3z(x)$ with one basis solution a power series expansion for $z$, and the other possibly having logarithm terms (which turns out not to be the case here).

Answer 2

In comments, you wrote that

The professor said the correct answer shouldn't involve too much algebra and the solution will most likely be some known functions like series for ex,cos(x) or something like that....i hardly see my solution be one of those..

This is fully true. Before doing anything else, let $\color{red}{y=x^3\,z}$ to make the equation $$x^3(z''+z)=0$$ Now, everything is more than simple.

Answer 3

As of @JerryHolmes' request, I will show how I solved it.

There is no need to overcomplicate stuff by using Frobenius' method; we can seek a simple power-series solution $$y(x)=\sum_{n=0}^\infty a_nx^n.$$

If we substitute the series with its derivatives into ODE, you should get $$\sum_{n=0}^\infty(n-4)(n-3)a_nx^{n-2}+\sum_{n=0}^\infty a_nx^n=0.$$

Notice that to combine the $2$ series, we have to set the first $2$ terms of the lagging series equal to $0.$ If we do so, we get $a_0=a_1=0.$

Now we can combine the $2$ series if we increase the index of the lagging series by $2$ $$\sum_{n=0}^\infty((n-1)(n-2)a_{n+2}+a_n)x^n=0.$$ So we get $$(n-2)(n-1)a_{n+2}+a_n=0.\tag{$\star$}\label{1}$$ Notice that we are going to solve for the coefficients via the $a_{n+2}$ term, so we don't want its coefficient in $\eqref{1}$ to be $0$ because if that happens, we can't solve for them so we will have to assume them as initial conditions. This happens for $n=1,2\implies a_{n+2}=a_3,a_4.$

So we know (or delt with) $a_0,a_1,a_3,a_4.$ To solve for $a_2,$ plug $n=0$ into $\eqref{1}$ to get $a_2=0.$

Now, we need to solve for $a_{n>4}.$ We can now safely solve explicitly for $a_{n+2}$ without the risk of dividing by $0.$ $$a_{n+2}=-\frac{a_n}{(n-2)(n-1)}.$$ Since the coefficients jump by $2,$ an even coefficient will depend on the previous even one and the same follows for the odd coefficients. $$a_{2n+1}=-\frac{a_{2n-1}}{(2n-3)(2n-2)}\\ a_{2n}=-\frac{a_{2n-2}}{(2n-4)(2n-3)}$$ To avoid confusion in this part, set $2k+1=n+2$ to solve for the odd coefficients and set $2k=n+2$ to solve for the even coefficients.

Since we start with $a_3$ and $a_4,$ odd coefficients will depend on $a_3$ and even coefficients will depend on $a_4.$

If you solve for the first few coefficients (or just notice the fact that we just keep dividing by consecutive integers,) you will notice the pattern $$a_{2n+1}=(-1)^{n+1}\frac{a_3}{(2n-2)!}\\ a_{2n}=(-1)^{n+1}\frac{a_4}{(2n-3)!}.$$

Now we can write the general solution $$y(x)=a_3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}x^{2n+1}+\\a_4\sum_{n=2}^\infty\frac{(-1)^{n+1}}{(2n-3)!}x^{2n}.$$

If we notice that $a_3=\frac{y^{(3)}(0)}{3!}$ and $a_4=\frac{y^{(4)}(0)}{4!}$ because of Mclaurin series, we can write it as $$y(x)=\frac{y^{(3)}(0)}{3!}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}x^{2n+1}+\\\frac{y^{(4)}(0)}{4!}\sum_{n=2}^\infty\frac{(-1)^{n+1}}{(2n-3)!}x^{2n}.$$ We can factor $x^3$ and increase the index of the first series by $1$ and that of the second series by $2$ to get $$y(x)=x^3\left(\frac{y^{(3)}(0)}{3!}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}-\\\frac{y^{(4)}(0)}{4!}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}\right)$$ which can be written in terms of familiar functions $$y(x)=x^3\left(\frac{y^{(3)}(0)}{3!}\cos x-\\\frac{y^{(4)}(0)}{4!}\sin x\right)$$ as your professor said.