Chapter 11.7, Problem 1AP

For Problems $1-13$ determine whether $x=0$ is an ordinary point or a regular singular point of the given differential equation. Then obtain two linearly independent solutions to the differential equation and state the maximum interval on which your solutions are valid.

$y^{″}+xy=0$

To determine

Whether $x=0$ is an ordinary point or a regular singular point of the given differential equation and obtain two linearly independent solutions to the differential equation and state the maximum interval on which your solutions are valid.

Answer to Problem 1AP

Solution:

The point $x=0$ is an ordinary point and the two linearly independent solutions are,

$\begin{array}{l}{y}_1\left(x\right)=1+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-2\right)\left(3n-5\right),\mathrm{......7}\cdot 4\cdot 1}{3n!}}{x}^{3n}\\ y_2\left(x\right)=x+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-1\right)\left(3n-4\right),\mathrm{......5}\cdot 2}{\left(3n+1\right)!}}{x}^{3n+1}\end{array}$

valid on $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

The differential equation is,

$y^{″}+xy=0$.

Approach:

The point $x=x_0$ is called an ordinary point of the differential equation $y^{″}+p\left(x\right)y^{\prime }+q\left(x\right)y=0$, if $p$ and $q$ are analytic at $x=x_0$.

Suppose their power series expansions are valid for $\left|x-x_0\right|<R$ then the general solution to the differential equation $y^{″}+p\left(x\right)y^{\prime }+q\left(x\right)y=0$ can be represented as a power series centered at $x=x_0$ with radius of convergence $R$.

The ratio test is,

$\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=l$,

and if $l<1$ then the series is convergent.

Calculation:

The point $x=x_0$ is called an ordinary point of the differential equation $y^{″}+p\left(x\right)y^{\prime }+q\left(x\right)y=0$, if $p$ and $q$ are analytic at $x=x_0$.

In the given equation $y^{″}+xy=0$,

$p\left(x\right)=0$ and $q\left(x\right)=x$, both are analytic at $x=0$. So it is an ordinary point.

Let the general solution of the differential equation is

$y\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}\mathrm{...}\left(1\right)$

Differentiate equation $\left(1\right)$ with respect to $x$.

$\begin{array}{rll}{y}^{\prime }\left(x\right)& =& {\displaystyle \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}}\\ y^{″}\left(x\right)& =& {\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)a_n{x}^{n-2}}\end{array}$

Substitute ${\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)a_n{x}^{n-2}}$ for $y^{″}$ and ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ for $y$ in the given equation.

${\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)a_n{x}^{n-2}}+x{\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}=0$ and,

${\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)a_n{x}^{n-2}}+{\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^{n+1}}=0$

Replace $n$ by $k+2$ in the first summation and replace $n$ by k-1 in the second summation.

${\displaystyle \sum _{k=0}^{\infty }\left(k+2\right)\left(k+1\right)a_{k+2}{x}^k+{\displaystyle \sum _{k=1}^{\infty }{a}_{k-1}{x}^k=0}}$

Separate out the terms corresponding to $k=0$ and $k\ge 1$. $2a_2+{\displaystyle \sum _{k=1}^{\infty }\left[\left(k+2\right)\left(k+1\right)a_{k+2}+a_{k-1}\right]}{x}^k=0$

For $k=0$ and $k\ge 1$, it is obtained that $2a_2=0$.

Thus, $a_2=0$ and the general relation is,

$\begin{array}{rll}\left(k+2\right)\left(k+1\right)a_{k+2}+a_{k-1}& =& 0\\ \left(k+2\right)\left(k+1\right)a_{k+2}& =& -a_{k-1}\\ a_{k+2}& =& -\frac{1}{\left(k+2\right)\left(k+1\right)}{a}_{k-1},k& =& 1,2,3,\mathrm{...}\end{array}$

Use this relation to determine the appropriate values of the coefficients.

Substitute successively for $k$ into above relation and obtain the coefficient.

When $k=1$,

$a_3=-\frac{1}{3\cdot 2}{a}_0$

When $k=2$,

$a_4=-\frac{1}{4\cdot 3}{a}_1$

When $k=3$,

$\begin{array}{rll}{a}_5& =& -\frac{1}{5\cdot 4}{a}_2\\ & =& 0\end{array}$

When $k=4$,

$\begin{array}{rll}{a}_6& =& -\frac{1}{6\cdot 5}{a}_1\\ & =& \frac{1}{6\cdot 5\cdot 3\cdot 2}{a}_0\end{array}$

When $k=5$,

$\begin{array}{rll}{a}_7& =& -\frac{1}{7\cdot 6}{a}_4\\ & =& \frac{1}{7\cdot 6\cdot 4\cdot 3}{a}_1\end{array}$

Continue further to obtain $a_{3n}$.

$a_{3n}=\frac{{\left(-1\right)}^n}{3n\left(3n-1\right)\left(3n-3\right)\left(3n-4\right)\cdot \mathrm{...3}\cdot 2}{a}_0\mathrm{...}\left(2\right)$

and

$a_{3n+1}=\frac{{\left(-1\right)}^n}{\left(3n+1\right)\left(3n\right)\left(3n-2\right)\left(3n-3\right)\cdot \mathrm{...4}\cdot 3}{a}_1\mathrm{...}\left(3\right)$

Use equations $\left(3\right)$ and $\left(4\right)$ to show that for all values of $a_0$, $a_1$, a solution to the given differential equation is,

$\begin{array}{rll}y\left(x\right)& =& \left[\begin{array}{l}{a}_0\left(1-\frac{1}{3\cdot 2}{x}^3+\frac{1}{6\cdot 5\cdot 3\cdot 2}{x}^6-\frac{1}{9\cdot 8\cdot 6\cdot 5\cdot 3\cdot 2}{x}^9+\mathrm{.......}\right)\\ +a_1\left(x-\frac{1}{4\cdot 3}{x}^4+\frac{1}{7\cdot 6\cdot 4\cdot 3}{x}^7-\mathrm{......}\right)\end{array}\right]\\ & =& \left[\begin{array}{l}{a}_0\left[1+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{3n\left(3n-1\right)\left(3n-3\right)\left(3n-4\right)\cdot \mathrm{...3}\cdot 2}{x}^{3n}}\right]\\ +a_1\left[x+\frac{{\left(-1\right)}^n}{\left(3n+1\right)\left(3n\right)\left(3n-2\right)\left(3n-3\right)\cdot \mathrm{...4}\cdot 3}{x}^{3n+1}\right]\end{array}\right]\end{array}$

Substitute $1$ for $a_0$ and $0$ for $a_1$ in above equation. $\begin{array}{rll}{y}_1\left(x\right)& =& 1+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{3n\left(3n-2\right)\left(3n-3\right)\left(3n-4\right)\cdot \mathrm{...3}\cdot 2}}{x}^{3n}\\ & =& 1+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-2\right)\left(3n-5\right),\mathrm{......7}\cdot 4\cdot 1}{3n!}}{x}^{3n}\end{array}$

Now, substitute $0$ for $a_0$ and $1$ for $a_1$.

$\begin{array}{rll}{y}_2\left(x\right)& =& x+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{\left(3n+1\right)\left(3n\right)\left(3n-2\right)\left(3n-3\right)\cdot \mathrm{...4}\cdot 3}}{x}^{3n+1}\\ & =& x+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-1\right)\left(3n-4\right),\mathrm{......5}\cdot 2}{\left(3n+1\right)!}}{x}^{3n+1}\end{array}$

For checking the solution is valid or not using the ratio test on both foregoing solution

Suppose,

$\begin{array}{rll}{a}_n& =& {\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-2\right)\left(3n-5\right),\mathrm{......7}\cdot 4\cdot 1}{3n!}}{x}^{3n}\\ a_{n+1}& =& {\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(3n+1\right)\left(3n-2\right),\mathrm{......7}\cdot 4\cdot 1}{3\left(n+1\right)!}}{x}^{3n+3}\end{array}$

Then by ratio test,

$\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=l$

And if $l<1$ then series is convergent.

So,

$\begin{array}{rll}\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}& =& \left[\underset{n\to \infty }{\lim }\left[\begin{array}{l}\left({\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(3n+1\right)\left(3n-2\right),\mathrm{......7}\cdot 4\cdot 1}{3\left(n+1\right)!}{x}^{3n+3}}\right)\\ *\left({\displaystyle \sum _{n=1}^{\infty }\frac{3n!}{{\left(-1\right)}^n\left(3n-2\right)\left(3n-5\right),\mathrm{......7}\cdot 4\cdot 1}{x}^{3n}}\right)\end{array}\right]\right]\\ & =& \underset{n\to \infty }{\lim }\frac{\left(-1\right)\left(3n+1\right)}{\left(3n+3\right)}\\ & =& \underset{n\to \infty }{\lim }\frac{\left(-1\right)n\left(3+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)}{n\left(3+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)}\\ & =& -\frac{1}{3}\end{array}$

From above $l<1$, so series is convergent.

Similarly for $y_2\left(x\right)$

$\begin{array}{rll}{a}_n& =& {\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-1\right)\left(3n-4\right),\mathrm{......5}\cdot 2}{\left(3n+1\right)!}}{x}^{3n+1}\\ a_{n+1}& =& {\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(3n+2\right)\left(3n-1\right),\mathrm{......5}\cdot 2}{\left(3n+4\right)!}}{x}^{3n+4}\end{array}$

Solve the above equations.

$\begin{array}{rll}\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}& =& \left[\underset{n\to \infty }{\lim }\left[\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(3n+2\right)\left(3n-1\right),\mathrm{......5}\cdot 2}{\left(3n+4\right)!}}{x}^{3n+4}\\ *{\displaystyle \sum _{n=1}^{\infty }\frac{\left(3n+1\right)}{\left({\left(-1\right)}^n\left(3n-1\right)\left(3n-4\right),\mathrm{......5}\cdot 2\right)x^{3n+1}}}\end{array}\right]\right]\\ & =& \underset{n\to \infty }{\lim }\frac{\left(-1\right)}{\left(3n+4\right)\left(3n+3\right)}\\ & =& 0\end{array}$

From above $l<1$. So, the series is convergent.

Therefore, both series are convergent and the solution are valid on $\left(-\infty ,\infty \right)$.

Conclusion:

Hence, the point $x=0$ is the ordinary point and the two independent solutions are,

$\begin{array}{l}{y}_1\left(x\right)=1+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-2\right)\left(3n-5\right),\mathrm{......7}\cdot 4\cdot 1}{3n!}}{x}^{3n}\\ y_2\left(x\right)=x+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(3n-1\right)\left(3n-4\right),\mathrm{......5}\cdot 2}{\left(3n+1\right)!}}{x}^{3n+1}\end{array}$

valid on $\left(-\infty ,\infty \right)$.