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6.2 Solutions About Ordinary Points: Class Notes

6.2a

Consider the differential equation $a_2(x)y″+a_1(x)y\prime +a_0(x)y=0$, which in standard form is $y″+P(x)y\prime +Q(x)y=0$.

Definition

A point $x=x_0$ is an ordinary point if $P$ and $Q$ are analytic at $x_0$ (meaning power series representations centered at $x=x_0$) exist. A point that is not ordinary is singular.

Theorem

If $x=x_0$ is an ordinary point of $a_2(x)y″+a_1(x)y\prime +a_0(x)y=0$, then we can always find two linearly independent solutions of the form $y=\sum _{n=0}^{\infty }{c}_n(x-x_0{)}^n$.

For us, $x_0=0$.

Example

Find two power series solutions of the given differential equation about the ordinary point $x=0$.

$$y″+2xy\prime +2y=0$$

Solution

Write $y$, $y\prime$, and $y″$ as series:

$$y=\sum _{n=0}^{\infty }{c}_n{x}^n,y\prime =\sum _{n=1}^{\infty }n{c}_n{x}^{n-1},y″=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}$$

Substitute into the equation:

$$\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+2x\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}+2\sum _{n=0}^{\infty }{c}_n{x}^n=0$$

Bring the constants and the $x$ inside the series:

$$\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+\sum _{n=1}^{\infty }2n{c}_n{x}^n+\sum _{n=0}^{\infty }2c_n{x}^n=0$$

Tip: Avoid negative indices: keep the $y″$ series starting at $n=2$ (its first two terms are zero), so after the shift it starts at $k=0$, not at a negative index.

Exponents. Two series already have $x^n$. Shift the $y″$ series to match: let $k=n-2$, so $n=k+2$. In the other two series, rename $n$ as $k$.

$$\sum _{k=0}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }2k{c}_k{x}^k+\sum _{k=0}^{\infty }2c_k{x}^k=0$$

Indices. The green series starts at $k=1$. Write out the $k=0$ terms of the other two series so that all three start at $k=1$:

$$2c_2+2c_0+\sum _{k=1}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }2k{c}_k{x}^k+\sum _{k=1}^{\infty }2c_k{x}^k=0$$

$$2c_2+2c_0+\sum _{k=1}^{\infty }[(k+2)(k+1)c_{k+2}+2k{c}_k+2c_k]x^k=0+0x+0x^2+\cdots$$

The right side is $0+0x+0x^2+\cdots$, so each coefficient on the left is zero. The constant terms give

$$2c_2+2c_0=0\Rightarrow c_2=-c_0$$

For $k\ge 1$, combine $2k{c}_k+2c_k=2(k+1)c_k$:

$$(k+2)(k+1)c_{k+2}+2(k+1)c_k=0,k=1,2,3,\dots$$

Divide by $k+1$ and solve for $c_{k+2}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_3}& {\displaystyle =-\frac{2}{3}{c}_1}\hfill \\ \hfill {\displaystyle k=2:c_4}& {\displaystyle =-\frac{2}{4}{c}_2=\frac{1}{2}{c}_0}\hfill \\ \hfill {\displaystyle k=3:c_5}& {\displaystyle =-\frac{2}{5}{c}_3=\frac{4}{15}{c}_1}\hfill \end{array}}$$

Substitute the coefficients into the series, then group the $c_0$ terms and the $c_1$ terms:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x-c_0{x}^2-\frac{2}{3}{c}_1{x}^3+\frac{1}{2}{c}_0{x}^4+\frac{4}{15}{c}_1{x}^5+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0(1-x^2+\frac{1}{2}{x}^4+\cdots )+c_1(x-\frac{2}{3}{x}^3+\frac{4}{15}{x}^5+\cdots )}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0{y}_1+c_1{y}_2}\hfill \end{array}}$$

6.2b

Example

Find two power series solutions of the given differential equation about the ordinary point $x=0$.

$$(x+2)y″+xy\prime -y=0$$

Solution

Plan: substitute the series, then match the exponents, match the indices, and find the recurrence relation.

$$y=\sum _{n=0}^{\infty }{c}_n{x}^n,y\prime =\sum _{n=1}^{\infty }n{c}_n{x}^{n-1},y″=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}$$

Substitute, distributing $(x+2)$ over the $y″$ series:

$$\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-1}+\sum _{n=2}^{\infty }2n(n-1)c_n{x}^{n-2}+\sum _{n=1}^{\infty }n{c}_n{x}^n-\sum _{n=0}^{\infty }{c}_n{x}^n=0$$

Exponents. Let $k=n-1$ in the first series (it starts at $k=1$), $k=n-2$ in the second (it starts at $k=0$), and $k=n$ in the other two:

$$\sum _{k=1}^{\infty }k(k+1)c_{k+1}{x}^k+\sum _{k=0}^{\infty }2(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }k{c}_k{x}^k-\sum _{k=0}^{\infty }{c}_k{x}^k=0$$

Indices. Write out the $k=0$ terms of the second and fourth series, so every series starts at $k=1$. Then combine, using $k{c}_k-c_k=(k-1)c_k$:

$$4c_2-c_0+\sum _{k=1}^{\infty }[k(k+1)c_{k+1}+2(k+2)(k+1)c_{k+2}+(k-1)c_k]x^k=0$$

Recurrence relation. Each coefficient is zero. The constant term gives $4c_2-c_0=0$, so ${\displaystyle c_2=\frac{1}{4}{c}_0}$. For $k\ge 1$,

$$k(k+1)c_{k+1}+2(k+2)(k+1)c_{k+2}+(k-1)c_k=0$$

Solve for $c_{k+2}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_3}& {\displaystyle =-\frac{1}{6}{c}_2-0=-\frac{1}{24}{c}_0}\hfill \\ \hfill {\displaystyle k=2:c_4}& {\displaystyle =-\frac{1}{4}{c}_3-\frac{1}{24}{c}_2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{96}{c}_0-\frac{1}{96}{c}_0=0}\hfill \\ \hfill {\displaystyle k=3:c_5}& {\displaystyle =-\frac{3}{10}{c}_4-\frac{1}{20}{c}_3=\frac{1}{480}{c}_0}\hfill \end{array}}$$

At $k=1$ the $c_1$ term drops out (its coefficient $k-1$ is 0), so every later coefficient is a multiple of $c_0$. The constant $c_1$ multiplies only $x$.

Two power series solutions: group the $c_0$ terms and the $c_1$ term.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+c_2{x}^2+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+\frac{1}{4}{c}_0{x}^2-\frac{1}{24}{c}_0{x}^3+\frac{1}{480}{c}_0{x}^5+\cdots }\hfill \end{array}}$$

6.2c

Example

Find two power series solutions of the given differential equation about the ordinary point $x=0$.

$$y″-(x+1)y\prime -y=0$$

Solution

Steps: exponents, indices, recurrence relation.

$$y=\sum _{n=0}^{\infty }{c}_n{x}^n,y\prime =\sum _{n=1}^{\infty }n{c}_n{x}^{n-1},y″=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}$$

Substitute, distributing $-(x+1)$ over the $y\prime$ series:

$$\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}-\sum _{n=1}^{\infty }n{c}_n{x}^n-\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0$$

Exponents. Let $k=n-2$ in the first series, $k=n-1$ in the third, and $k=n$ in the second and fourth:

$$\sum _{k=0}^{\infty }(k+1)(k+2)c_{k+2}{x}^k-\sum _{k=1}^{\infty }k{c}_k{x}^k-\sum _{k=0}^{\infty }(k+1)c_{k+1}{x}^k-\sum _{k=0}^{\infty }{c}_k{x}^k=0$$

Indices. Write out the $k=0$ terms of the first, third, and fourth series, and combine $-k{c}_k-c_k=-(k+1)c_k$:

$$2c_2-c_1-c_0+\sum _{k=1}^{\infty }[(k+1)(k+2)c_{k+2}-(k+1)c_{k+1}-(k+1)c_k]x^k=0$$

Recurrence relation. The constant term gives $2c_2-c_1-c_0=0$. For $k\ge 1$, every term in the bracket has a factor of $k+1$; divide it out and solve for $c_{k+2}$:

$${\displaystyle c_2=\frac{1}{2}{c}_1+\frac{1}{2}{c}_0,c_{k+2}=\frac{1}{k+2}{c}_{k+1}+\frac{1}{k+2}{c}_k,k=1,2,3,\dots }$$

Each coefficient involves both $c_0$ and $c_1$ (for example, ${\displaystyle c_3=\frac{1}{6}{c}_0+\frac{1}{2}{c}_1}$), so find the two solutions separately: first let $c_1=0$, then let $c_0=0$.

Let $c_1=0$ (so ${\displaystyle c_2=\frac{1}{2}{c}_0}$):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_3}& {\displaystyle =\frac{1}{3}{c}_2+\frac{1}{3}{c}_1=\frac{1}{6}{c}_0}\hfill \\ \hfill {\displaystyle k=2:c_4}& {\displaystyle =\frac{1}{4}{c}_3+\frac{1}{4}{c}_2=\frac{1}{24}{c}_0+\frac{1}{8}{c}_0=\frac{1}{6}{c}_0}\hfill \end{array}}$$

Let $c_0=0$ (so ${\displaystyle c_2=\frac{1}{2}{c}_1}$):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_3}& {\displaystyle =\frac{1}{3}{c}_2+\frac{1}{3}{c}_1=\frac{1}{6}{c}_1+\frac{1}{3}{c}_1=\frac{1}{2}{c}_1}\hfill \\ \hfill {\displaystyle k=2:c_4}& {\displaystyle =\frac{1}{4}{c}_3+\frac{1}{4}{c}_2=\frac{1}{8}{c}_1+\frac{1}{8}{c}_1=\frac{1}{4}{c}_1}\hfill \end{array}}$$

Put the coefficients into the series:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0+\frac{1}{2}{c}_0{x}^2+\frac{1}{6}{c}_0{x}^3+\frac{1}{6}{c}_0{x}^4+\cdots }\hfill \\ \hfill {\displaystyle }& {\displaystyle +c_{1}x+\frac{1}{2}{c}_1{x}^2+\frac{1}{2}{c}_1{x}^3+\frac{1}{4}{c}_1{x}^4+\cdots }\hfill \end{array}}$$

Suppose we have initial conditions $y(0)=a,y\prime (0)=b$. Since $y(0)=c_0$ and $y\prime (0)=c_1$, this gives

$$c_0=a,c_1=b$$

6.2d

Use an appropriate Maclaurin series to find two power series solutions of the given differential equation about the ordinary point $x=0$.

$$y″+e^{x}y\prime -y=0$$

Solution

Replace $e^x$ with its Maclaurin series and substitute the series for $y$, $y\prime$, and $y″$:

$${\displaystyle \sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+(1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\cdots )\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0}$$

Write the $y\prime$ series out and multiply each term of the $e^x$ series by every term of it:

$${\displaystyle (1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\cdots )(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+\cdots )}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 1\text{ times}:}& {\displaystyle c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+\cdots }\hfill \\ \hfill {\displaystyle x\text{ times}:}& {\displaystyle c_{1}x+2c_2{x}^2+3c_3{x}^3+4c_4{x}^4+\cdots }\hfill \\ \hfill {\displaystyle \frac{1}{2}{x}^2\text{ times}:}& {\displaystyle \frac{1}{2}{c}_1{x}^2+c_2{x}^3+\frac{3}{2}{c}_3{x}^4+2c_4{x}^5+\cdots }\hfill \\ \hfill {\displaystyle \frac{1}{6}{x}^3\text{ times}:}& {\displaystyle \frac{1}{6}{c}_1{x}^3+\frac{1}{3}{c}_2{x}^4+\frac{1}{2}{c}_3{x}^5+\cdots }\hfill \end{array}}$$

Write out $y″$ and $y$ too. The whole left side is the sum of these rows:

$$\begin{array}{cc}\hfill {\displaystyle y″:}& {\displaystyle 2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+\cdots }\hfill \\ \hfill {\displaystyle e^{x}y\prime :}& {\displaystyle (\text{the four rows above})}\hfill \\ \hfill {\displaystyle -y:}& {\displaystyle -c_0-c_{1}x-c_2{x}^2-c_3{x}^3-\cdots }\hfill \end{array}$$

The sum equals 0, so the coefficient of each power of $x$ is zero. Constant terms:

$${\displaystyle 2c_2+c_1-c_0=0\Rightarrow c_2=-\frac{1}{2}{c}_1+\frac{1}{2}{c}_0}$$

$x$ terms (the $c_{1}x$ and $-c_{1}x$ cancel):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 6c_3+2c_2=0\Rightarrow c_3}& {\displaystyle =-\frac{1}{3}{c}_2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{6}{c}_1-\frac{1}{6}{c}_0}\hfill \end{array}}$$

Substitute into the series and group the $c_0$ terms and the $c_1$ terms:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+(-\frac{1}{2}{c}_1+\frac{1}{2}{c}_0)x^2+(\frac{1}{6}{c}_1-\frac{1}{6}{c}_0)x^3+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0(1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\cdots )+c_1(x-\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\cdots )}\hfill \end{array}}$$