# Determining the first $3$ nonzero terms in each of two linearly independent solutions

In preparation for an upcoming exam I have came across a question that I am a little confused with.

Given the following different equation,

$$xy''+y'-y = 0$$

I am trying to find the first $3$ nonzero terms in each of two linearly independent solutions.

I have determined the indicial equation to be,

$$r^2 = 0$$

Therefore the roots are,

$$r_1=0, r_2=0$$

So the roots are separated by a integer

I have calculated the recurrence solution to be,

$$a_n = \frac{a_{n-1}}{(n+r)^2}$$

So therefore the first solution would be the following,

$$y_1(x) = 1+ x + \frac{1}{4}x^2 + \frac{1}{36}x^3...$$

However it is determining the second solution where I am having problem, so I am looking for some help with showing how this answer is calculated, thanks! The solution to the second solution is,

$$y_2(x) = y_1(x)\ln x -2x - \frac{3}{4}x^2-\frac{11}{108}x^3$$

## 1 Answer

Since $r_1=r_2=0$, a second solution $y_2(x)$ has the form $$y_2(x) = y_1(x)\ln x +\sum_{n=1}^{\infty}b_nx^n.$$ We have $$y_2'(x)=y_1'\ln(x)+y_1x^{-1}+\sum_{n=1}^{\infty}nb_nx^{n-1}\\ y_2''(x)=y_1''\ln(x)+2x^{-1}y_1'-y_1x^{-2}+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-2}.$$ Inserting this in the equation we get $$2y_1'+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-1}+\sum_{n=1}^{\infty}nb_nx^{n-1}-\sum_{n=1}^{\infty}b_nx^n=0,$$ or equivalently, we write $$2a_1+2\sum_{n=2}^{\infty}na_nx^{n-1}+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-1}+b_1+\sum_{n=2}^{\infty}nb_nx^{n-1}-\sum_{n=2}^{\infty}b_{n-1}x^{n-1}=0.$$ Thus $$2a_1+b_1+\sum_{n=2}^{\infty}(2na_n+n^2n_n-b_{n-1})x^{n-1}=0.$$ By identification, we obtain $$\left\{\begin{array}{l} b_1=-2a_1\\ b_n=\frac{b_{n-1}-2na_n}{n^2},\quad n\geq2. \end{array}\right. \qquad\Rightarrow\left\{\begin{array}{l} b_1=-2\\ b_2=-\dfrac{3}{4}\\ b_3=-\dfrac{11}{108} \end{array}\right.$$

• do you choose a1 to be 1? – jh123 Dec 13 '17 at 21:02
• Actually, I took $a_0=y_1(0)=1$, so $y_1(x)=a_0+a_0x+\frac{a_0}{4}x^2+\frac{a_0}{36}x^3+...=a_0(1+x+\frac{1}{4}x^2+\frac{1}{36}x^3+...)$ – Wang Dec 13 '17 at 21:10
• when you say inserting into the equation, why does everything cancel? where did the $ln$ go? – jh123 Dec 13 '17 at 21:17
• did you choose x to be 1? – jh123 Dec 13 '17 at 21:18
• $y_1'(x)=\left(a_0+a_1x+\sum_{n=2}^{\infty}a_nx^{n}\right)'=a_1+\sum_{n=2}^{\infty}na_nx^{n-1}$ – Wang Dec 13 '17 at 21:40