Given the follow DE,

$$xy''+y = 0 $$

I am trying to determine the first 4 terms in each of the two linearly independent solutions about $x=0$

So I have calculated the indicial equation, indicial roots, recurrence equation, and the first solution. The roots differ by a integer so I am trying to use the formula to determine the second solution,

$$y_2(x) = y_1(x)\int\frac{W}{y_1^2}$$

where W is the coefficient of the $y'$

So here are my workings, $r=0,1$ and the recurrence formula is $a_n = -\frac{a_{n-1}}{(n+r)(n+r-1)}$

I have calculated the first solution to be using $r=1$,

$$y_1(x) = x - \frac{1}{2}x^2+ \frac{1}{12}x^3-\frac{1}{144}x^4$$

Here is my attempt at the second solution which is $y_2(x) = -y_1(x)\ln x + 1 -\frac{3}{4}x^2+\frac{7}{36}x^3-\frac{35}{1728}x^4$ and I know that $W=1$ since $e^{\int 0 dx } = 1$

$$y_2(x) = y_1(x)\int\frac{W}{y_1^2} \\= x - \frac{1}{2}x^2+ \frac{1}{12}x^3-\frac{1}{144}x^4 \int \frac{1}{(x - \frac{1}{2}x^2+ \frac{1}{12}x^3-\frac{1}{144}x^4)^2}$$

So how do I simplify that to get,

$$y_2(x) = -y_1(x)\ln x + 1 -\frac{3}{4}x^2+\frac{7}{36}x^3-\frac{35}{1728}x^4$$

  • $\begingroup$ @Moo I watched both of these videos before but I am trying to solve it through the method above, could you help? $\endgroup$ – jh123 Dec 13 '17 at 23:10
  • $\begingroup$ @moo I thought it was in the numerator now I am confused $\endgroup$ – jh123 Dec 13 '17 at 23:17
  • $\begingroup$ I will try and follow it $\endgroup$ – jh123 Dec 13 '17 at 23:21
  • $\begingroup$ Well there is no $y'$ term so shouldn't it be $0$ $\endgroup$ – jh123 Dec 13 '17 at 23:39

You get $$ \frac1{y_1^2}=\frac1{x^2(1+a_1x+a_2x^2+...)^2}=\frac1{x^2}(1+c_1+c_2x^2+c_3^3+\dots)=\frac1{x^2}+\frac{c_1}x+c_2+c_3x^2+... $$ Integrating this gives $$ \int\frac1{y_1^2}dx =-\frac1x+c_1\ln x+c_2x+\frac{c_3}2 x^2+... $$ so that after multiplication with $y_1$ you get one term $c_1y_1(x)\ln x$ plus some power series$v(x)=b_1x+b_2x^2+...$, as the leading term $-\frac1xy_1(x)=-1-a_1x-a_2x^2+...$ is also just a power series.

What exactly the coefficient of $y_1(x)\ln x$ is is arbitrary, as any non-trivial multiple of $y_2$ can as well serve as second basis solution. Using the integral as given, one should get $c_1=-2a_2=1$. If one wants $y_2(0)=1$, then one has to take the negative of the thus computed solution.


Plugging $y_2(x) = - y_1(x) \ln(x) + v(x)$ in to the equation should give you $$ \eqalign{x v'' + v &= - 2 y_1' - y_1'' \cr &= -1 + \frac{3}{2} x - \frac{5}{12} x^2 + \frac{7}{144} x^3 + \ldots}$$ If $v(x) = b_0 + b_1 x + b_2 x^2 + \ldots$, equating powers of $x$ should give you $$ \eqalign{b_0 &= -1\cr 2 b_2 + b_1 &= \frac{3}{2} \cr 6 b_3 + b_2 &= -\frac{5}{12}\cr 12 b_4 + b_3 &= \frac{7}{144} \cr \ldots &\cr}$$ $b_1$ is arbitrary, and may be taken to be $0$. Then you should get $b_2 = 3/4$, $b_3 =(-5/12-b_2)/6 = 7/36$, etc.

  • $\begingroup$ is it always a -y1(x)? on your first line of your solution $\endgroup$ – jh123 Dec 13 '17 at 23:46

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