1. 1. Assume [latex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n[/latex] (step 1). Then, [latex]y^\prime(x)=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}[/latex] and [latex]y^{\prime\prime}(x)=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/latex] (step 2). We want to find values for the coefficients [latex]a_n[/latex] such that

[latex]\begin{aligned} y^{\prime\prime}-y&=0 \\ \displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\displaystyle\sum_{n=0}^\infty a_nx^n&=0 \end{aligned}[/latex] (step 3).

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with [latex]n=0[/latex]. 
To re-index the first term, replace [latex]n[/latex] with [latex]n+2[/latex] inside the sum, and change the lower summation limit to [latex]n=0[/latex]. We get

[latex]\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/latex].

This gives

[latex]\begin{aligned} \displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\displaystyle\sum_{n=0}^\infty a_nx^n&=0 \\ \displaystyle\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-a_n]x^n&=0 \text{ (step 4)}. \end{aligned}[/latex]

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of [latex]x[/latex] are zero. So we have

[latex]\large{(n+2)(n+1)a_{n+2}-a_n=0\text{ for }n=0,1,2,\ldots}[/latex]

This recurrence relationship allows us to express each coefficient [latex]a_n[/latex] in terms of the coefficient two terms earlier. This yields one expression for even values of [latex]n[/latex] and another expression for odd values of [latex]n[/latex]. Looking first at the equations involving even values of [latex]n[/latex], we see that

[latex]\begin{aligned} a_2&=\frac{a_0}2 \\ a_4&=\frac{a_2}{4\cdot3}=\frac{a_0}{4!} \\ a_6&=\frac{a_4}{6\cdot5}=\frac{a_0}{6!} \\ &\vdots \end{aligned}[/latex]

Thus, in general, when [latex]n[/latex] is even, [latex]a_n=\frac{a_0}{n!}[/latex] (step 5).
For the equations involving odd values of [latex]n[/latex], we see that

[latex]\begin{aligned} a_3&=\frac{a_1}{3\cdot2}=\frac{a_1}{3!} \\ a_5&=\frac{a_3}{5\cdot4}=\frac{a_1}{5!} \\ a_7&=\frac{a_5}{7\cdot6}=\frac{a_1}{7!} \\ &\vdots \end{aligned}[/latex]

Therefore, in general, when [latex]n[/latex] is odd, [latex]a_n=\frac{a_1}{n!}[/latex] (step 5 continued).
Putting this together, we have

[latex]\begin{aligned} y(x)&=\displaystyle\sum_{n=0}^\infty a_nx^n \\ &=a_0+a_1x+\frac{a_0}2x^2+\frac{a_1}{3!}x^3+\frac{a_0}{4!}x^4+\frac{a_1}{5!}x^5+\cdots \end{aligned}[/latex].

Re-indexing the sums to account for the even and odd values of [latex]n[/latex] separately, we obtain

[latex]y(x)=a_0\displaystyle\sum_{k=0}^\infty\frac{1}{(2k)!}x^{2k}+a_1\displaystyle\sum_{k=0}^\infty\frac{1}{(2k+1)!}x^{2k+1}[/latex].

Analysis for part a.
As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since [latex]y(x)=c_1e^x+c_2e^{-x}[/latex] is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
Fortunately, after writing the power series representations of [latex]e^x[/latex] and [latex]e^{-x}[/latex], and doing some algebra, we find that if we choose

[latex]\large{c_0=\frac{(a_0+a_1)}2, \ c_1=\frac{(a_0-a_1)}2}[/latex],

we then have [latex]a_0=c_0+c_1[/latex] and [latex]a_1=c_0-c_1[/latex], and

[latex]\begin{aligned} y(x)&=a_0+a_1x+\frac{a_0}2x^2+\frac{a_1}{3!}x^3+\frac{a_0}{4!}x^4+\frac{a_0}{5!}x^5+\ldots \\ &=(c_0+c_1)+(c_0-c_1)x+\frac{(c_0+c_1)}2x^2+\frac{(c_0-c_1)}{3!}x^3+\frac{(c_0+c_1)}{4!}x^4+\frac{(c_0-c_1)}{5!}x^5+\ldots \\ &=c_0\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}+c_1\displaystyle\sum_{n=0}^\infty\frac{(-x)^n}{n!} \\ &=c_0e^x+c_1e^{-x} \end{aligned}[/latex].

So we have, in fact, found the same general solution. Note that this choice of [latex]c_1[/latex] and [latex]c_2[/latex] is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.

• Assume [latex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n[/latex] (step 1). Then, [latex]y^\prime(x)=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}[/latex] and [latex]y^{\prime\prime}=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/latex] (step 2). We want to find values for the coefficients [latex]a_n[/latex] such that

[latex]\begin{aligned} (x^2-1)y^{\prime\prime}+6xy^\prime+4y&=-4 \\ (x^2-1)\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+6x\displaystyle\sum_{n=1}^\infty na_nx^{n-1}+4\displaystyle\sum_{n=0}^\infty a_nx^n&=-4 \\ x^2\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+6x\displaystyle\sum_{n=1}^\infty na_nx^{n-1}+4\displaystyle\sum_{n=0}^\infty a_nx^n&=-4 \end{aligned}[/latex].

Taking the external factors inside the summations, we get

[latex]\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n}-\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\displaystyle\sum_{n=1}^\infty 6na_nx^n+\displaystyle\sum_{n=0}^\infty 4a_nx^n=-4[/latex] (step 3).

Now, in the first summation, we see that when [latex]n=0[/latex] or[latex]n=1[/latex], the term evaluates to zero, so we can add these terms back into our sum to get

[latex]\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n}=\displaystyle\sum_{n=0}^\infty n(n-1)a_nx^{n}[/latex].

Similarly, in the third term, we see that when [latex]n=0[/latex], the expression evaluates to zero, so we can add that term back in as well. We have

[latex]\displaystyle\sum_{n=1}^\infty 6na_nx^n=\displaystyle\sum_{n=0}^\infty 6na_nx^n[/latex].

Then, we need only shift the indices in our second term. We get

[latex]\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/latex]

Thus, we have

[latex]\begin{aligned} \displaystyle\sum_{n=0}^\infty n(n-1)a_nx^{n}-\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n+\displaystyle\sum_{n=0}^\infty6na_nx^n+\displaystyle\sum_{n=0}^\infty4a_nx^n&=-4\text{ (step 4).} \\ \displaystyle\sum_{n=0}^\infty[n(n-1)a_n-(n+2)(n+1)a_{n+2}6na_n+4a_n]x^n&=-4 \\ \displaystyle\sum_{n=0}^\infty[(n^2-n)a_n+6na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\ \displaystyle\sum_{n=0}^\infty[n^2a_n+5na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\ \displaystyle\sum_{n=0}^\infty[(n^2+5n+4)a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\ \displaystyle\sum_{n=0}^\infty[(n+4)(n+1)a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \end{aligned}[/latex]

Looking at the coefficients of each power of [latex]x[/latex], we see that the constant term must be equal to [latex]-4[/latex] and the coefficients of all other powers of [latex]x[/latex] must be zero. Then, looking first at the constant term,

[latex]\begin{aligned} 4a_0-2a_2&=-4 \\ a_2&=2a_0+2 \end{aligned}[/latex] (step 3).

For [latex]n\geq1[/latex], we have

[latex]\begin{aligned} (n+4)(n+1)a_n-(n+2)(n+1)a_{a+2}&=0 \\ (n+1)[(n+4)a_n-(n+2)a_{n+2}]&=0 \end{aligned}[/latex].

Since [latex]n\geq1[/latex], [latex]n+1\ne0[/latex], we see that

[latex](n+4)a_n-(n+2)a_{n+2}=0[/latex].

and thus

[latex]a_{n+2}=\frac{n+4}{n+2}a_n[/latex].

For even values of [latex]n[/latex], we have

[latex]\begin{aligned} a_4&=\frac64(2a_0+2)=3a_0+3 \\ a_6&=\frac86(3a_0+3)=4a_0+4 \\ &\vdots \end{aligned}[/latex].

In general, [latex]a_{2k}=(k+1)(a_0+1)[/latex] (step 5).
For odd values of [latex]n[/latex], we have

[latex]\begin{aligned} a_3&=\frac53a_1 \\ a_5&=\frac75a_3=\frac73a_1 \\ a_7&=\frac97a_5=\frac93a_1=3a_1 \\ &\vdots \end{aligned}[/latex].

In general, [latex]a_{2k+1}=\frac{2k+3}3a_1[/latex] (step 5 continued).
Putting this together, we have

[latex]y(x)=\displaystyle\sum_{k=0}^\infty (k+1)(a_0+1)x^{2k}+\displaystyle\sum_{k=0}^\infty\left(\frac{2k+3}3\right)a_1x^{2k+1}[/latex] (step 6).

The Bessel equation of order [latex]0[/latex] is given by

[latex]x^2y^{\prime\prime}xy^\prime+x^2y=0[/latex].

We assume a solution of the form [latex]y=\displaystyle\sum_{n=0}^\infty a_nx^n[/latex]. Then [latex]y^\prime(x)=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}[/latex] and [latex]y^{\prime\prime}(x)=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/latex]. Substituting this into the differential equation, we get

[latex]\begin{aligned} x^2\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+x\displaystyle\sum_{n=1}^\infty na_nx^{n-1}+x^2\displaystyle\sum_{n=0}^\infty a_nx^n&=0 &\quad &\text{Substitution} \\ \displaystyle\sum_{n=2}^\infty n(n-1)a_nx^n+\displaystyle\sum_{n=1}^\infty na_nx^n+\displaystyle\sum_{n=0}^\infty a_nx^{n+2} &=0 &\quad &\text{Bring external factors within sums} \\ \displaystyle\sum_{n=2}^\infty n(n-1)a_nx^n+\displaystyle\sum_{n=1}^\infty na_nx^n+\displaystyle\sum_{n=2}^\infty a_{n-2}x^n&=0 &\quad &\text{Re-index third sum.} \\ \displaystyle\sum_{n=2}^\infty n(n-1)a_nx^n+a_1x+\displaystyle\sum_{n=2}^\infty na_nx^n+\displaystyle\sum_{n=2}^\infty a_{n-2}x^n&=0 &\quad &\text{Separate }n=1\text{ term from second sum} \\ a_1x+\displaystyle\sum_{n=2}^\infty [n(n-1)a_n+na_n+a_{n-2}]x^n&=0 &\quad &\text{Collect summation terms.} \\ a_1x+\displaystyle\sum_{n=2}^\infty[(n^2-n)a_n+na_n+a_{n-2}]x^n&=0 &\quad &\text{Multiply through first term.} \\ a_1x+\displaystyle\sum_{n=2}^\infty [n^2a_n+a_{n-2}]x^n&=0 &\quad &\text{Simplify.} \end{aligned}[/latex]

Then [latex]a_1=0[/latex], and for [latex]n\geq2[/latex],

[latex]\begin{aligned} n^2a_n+a_{n-2}&=0 \\ a_n=-\frac{1}{n^2}a_{n-2} \end{aligned}[/latex].

Because [latex]a_1=0[/latex], all odd terms are zero. Then, for even values of [latex]n[/latex], we have

[latex]\begin{aligned} a_2&=-\frac{1}{2^2}a_0 \\ a_4&=-\frac{1}{4^2}a_2=\frac{1}{4^2\cdot2^2}a_0 \\ a_6&=-\frac{1}{6^2}a_4=-\frac{1}{6^2\cdot4^2\cdot2^2}a_0 \end{aligned}[/latex].

In general,

[latex]a_{2k}=\frac{(-1)^k}{(2)^{2k}(k!)^2}a_0[/latex].

Thus, we have

[latex]y(x)=a_0\displaystyle\sum_{k=0}^\infty\frac{(-1)^k}{(2)^{2k}(k!)^2}x^{2k}[/latex].

The graph appears below.

Figure 1.