Series Solution(case of ordinary point)

Exercise4.2

1. Find the series solution of the following differential equations about $x = 0$.

(a) $\ y^{\prime} + y = e^{x}$
(b) $\ y^{\prime\prime} + xy = 0$
(c) $\ xy^{\prime} - y = x^{2}e^{x}$
(d) $\ y^{\prime\prime} + xy^{\prime} - 4y = 0$
2. Find the series solution of the following differential equations about the indicated point.

(a) $\ y^{\prime\prime} + (x-2)y = 0, \ a = 2$
(b) $\ y^{\prime\prime} + xy^{\prime} + 3y = x^2, \ a = 1$

4.2
1.
(a) Since $P(x) = 1, Q(x) = 1, R(x) = e^{x}$，$x = 0$ is an ordinary point. Thus, we set the solution

$$y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$$

Then

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

Since $e^{x} = \sum_{n=0}^{\infty}x^{n}/n!$, we substitute this into the given equation. Then

$$\sum_{n=1}^{\infty}nc_{n}x^{n-1} + \sum_{n=0}^{\infty}c_{n}x^{n} = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$

Simplifying this,

$$\sum_{n=1}^{\infty}nc_{n}x^{n-1} + \sum_{n=0}^{\infty}(c_{n} - \frac{1}{n!})x^{n} = 0$$

Now align the power of $x$ to be the least power $n-1$.

$$\sum_{n=1}^{\infty}[nc_{n} + c_{n-1} - \frac{1}{(n-1)!}]x^{n-1} = 0 .$$

Using term by term differentiation. the coefficients of $x^{n-1}$ are all 0. Thus, we have

$$nc_{n} + c_{n-1} - \frac{1}{(n-1)!} = 0 , n \geq 1 .$$

or

$$c_{n} = \frac{1}{n!} - \frac{c_{n-1}}{n}$$

Now that the value of $c_{0}$ is determined by $y(0)$. In this case, $c_0$ is treated as constant. Thus, we find $c_{1},c_{2},\ldots$. Then

$$c_{1} = 1 - c_{0}, c_{2} = \frac{c_{0}}{2}, c_{3} = \frac{1}{3!} - \frac{c_{0}}{3!}, \cdots$$

From this,

$$c_{2n} = \frac{c_{0}}{(2n)!)}, c_{2n+1} = \frac{1 - c_{0}}{(2n+1)!} .$$

Put this into $y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$. Then

$$y = \sum_{n=0}^{\infty}c_{2n}x^{2n} + \sum_{n=0}^{\infty}c_{2n+1}x^{2n+1}$$

$$= c_{0}\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} - c_{0} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$

$$= c_{0}\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{n}}{(n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$

Therefore, the solution is given by the following function.

$$y = c_{0}e^{-x} + \frac{e^{x}- e^{-x}}{2}.$$

(b) Since $P(x) = 0, Q(x) = x, R(x) = 0$，$x = 0$ is an ordinary point. Then we set our solution as

$$y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$$

Then

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

$$y^{\prime\prime} = \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}$$

Put these into the given equation. We have

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=0}^{\infty}c_{n}x^{n+1} = 0$$

Now align the power of $x$ to be the smallest $n-2$. Then

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=3}^{\infty} c_{n-3}x^{n-2} = 0 .$$

Next align the series index to the largest one.

$$2c_{2} + \sum_{n=3}^{\infty}[n(n-1)c_{n} + c_{n-3}]x^{n-2} = 0$$

Using the term by term differentiation, the coefficients of $x^{n-1}$are all 0. Then we have the following recurrence relation.

$$c_{2} = 0, n(n-1)c_{n} + c_{n-3} = 0 , n \geq 3 .$$

or

$$c_{2} = 0, c_{n} = -\frac{c_{n-3}}{n(n-1)}$$

Note that the value of $c_{0},c_{1}$ are determined by $y(0), y^{\prime}(0)$. So, we can think of them as constants. Thus, we find $c_{2},c_{3},\ldots$ interms of $c_0, c_1$.

$$c_{3} = \frac{c_{0}}{3 \cdot 2}, c_{4} = \frac{-c_{1}}{4 \cdot 3}, c_{5} = 0, \cdots$$

Thus

$$c_{3n} = \frac{c_{3n}}{c_{3n-3}}\frac{c_{3n-3}}{c_{3n-6}}\cdots\f... ...c_{3}}{c_{0}}c_{0} = \frac{(-1)^{n}(3n-2)(3n-5) \cdots 4 \cdot 1}{(3n)!}c_{0},$$

$$c_{3n+1} = \frac{c_{3n+1}}{c_{3n-2}}\frac{c_{3n-2}}{c_{3n-5}}\cdo... ...{4}}{c_{1}}c_{1} = \frac{(-1)^{n}(3n-1)(3n-4) \cdots 5 \cdot 2}{(3n+1)!}c_{1},$$

Put this into $y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$. Then

$$y = \sum_{n=0}^{\infty}c_{3n}x^{3n} + \sum_{n=0}^{\infty}c_{3n+1}x^{3n+1}$$

where

$$c_{3n} = \frac{(-1)^{n}(3n-2)(3n-5) \cdots 4\cdot1}{(3n)!}c_{0}, c_{3n+1} = \frac{(-1)^{n}(3n-1)(3n-4) \cdots 5\cdot2}{(3n+1)!}c_{1},$$

(c) Since $P(x) = x, Q(x) = -1, R(x) = x^2 e^x$，$x = 0$ is an ordinary point. Then we set

$$y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$$

Then，

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

Put this into the given equation.

$$\sum_{n=1}^{\infty}nc_{n}x^{n} - \sum_{n=0}^{\infty}c_{n}x^{n} = \sum_{n=0}^{\infty}\frac{x^{n+2}}{n!}$$

Now we align the power of $x$ to be the least $n$. Then

$$\sum_{n=1}^{\infty}nc_{n}x^{n} + \sum_{n=0}^{\infty}c_{n}x^{n} - \sum_{n=2}^{\infty}\frac{x^{n}}{(n+2)!} = 0 .$$

Next we align the series indexto the largest one. Then

$$c_{1}x - c_{0} - c_{1}x + \sum_{n=2}^{\infty}[nc_{n} - c_{n} - \frac{1}{(n-2)!}]x^{n} = 0.$$

Using term by term differentiation, the coefficients of $x^{n}$ are all 0. Thus, we have the recurrence relation.

$$c_{0} = 0, (n-1)c_{n} - \frac{1}{(n-2)!} = 0 , n \geq 2 .$$

or

$$c_{0} = 0, c_{n} = \frac{1}{(n-1)!}$$

Note that the value of $c_1$ is determined by the initial condition $y^{\prime}(0)$. Thus, we can think of $c_1$ as a constant.

$$y = \sum_{n=0}^{\infty}c_{n}x^{n}$$

$$= c_{1}x + \sum_{n=2}^{\infty} c_{n} x^{n}$$

$$= c_{1}x + \sum_{n=2}^{\infty} \frac{x^{n}}{(n-1)!}$$

$$= c_{1}x + x(\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} - 1)$$

Therefore, the solution is given by the following elementary function.

$$y = c_{1}x + x(e^{x} - 1).$$

(d) Since $P(x) = x, Q(x) = -4, R(x) = 0$，$x = 0$ is an ordinary point. Thus we set

$$y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$$

Then

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

$$y^{\prime\prime} = \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}$$

Now put these into the given equation.

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=0}^{\infty}nc_{n}x^{n} - 4\sum_{n=0}^{\infty}c_{n}x^{n} = 0$$

Then align the power of $x$ to the least $n$.

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=3}^{\infty}(n-2)c_{n-2}x^{n-2} - 4\sum_{n=2}^{\infty}c_{n-2}x^{n-2} = 0 .$$

We next align the series index to the largest one.

$$2c_{2} - 4c_{0} + \sum_{n=3}^{\infty}[n(n-1)c_{n} + (n-6)c_{n-2}]x^{n-2} = 0.$$

Using the term by term differentiation, we see that the coefficients of $x^{n}$ are all 0. So, we have the recurrent equation.

$$2c_{2} - 4c_{0} = 0, n(n-1)c_{n} + (n-6)c_{n-2} = 0 , n \geq 3$$

or

$$c_{2} = 2c_{0}, c_{n} = -\frac{(n-6)c_{n-2}}{n(n-1)}, n \geq 3$$

Note that $c_{0},c_{1}$ are determined by $y(0), y^{\prime}(0)$. Then we can think of these as arbitrary constants. Thus, we find $c_{2},c_{3},\ldots$ interms of $c_0, c_1$. Now

$$c_{2} = 2c_{0}, c_{3} = \frac{-c_{1}}{2}, \cdots$$

Thus,

$$c_{2n} = \frac{c_{2n}}{c_{2n-2}}\frac{c_{2n-2}}{c_{2n-4}}\frac{c_{2n-4}}{c_{2n-6}}c_{2n-6}, n \leq 2 c_{2n} = 0, n \geq 3$$

Also,

$$c_{2n+1} = \frac{c_{2n+1}}{c_{2n-1}}\frac{c_{2n-1}}{c_{2n-3}}\cdo... ...{c_{3}}{c_{1}}c_{1} = \frac{3(-1)^{n}c_{1}}{2^n n! (2n+1)(2n-1)(2n-3)}x^{2n+1}$$

Thus, we have

$$y = \sum_{n=0}^{\infty}c_{n}x^{n}$$

$$= c_{0} + 2c_{0}x^{2} - \frac{c_{0}}{3}x^4 - c_{1}\sum_{n=2}^{\infty} \frac{3(-1)^{n}}{2^n n! (2n+1)(2n-1)(2n-3)}x^{2n+1}$$

2.
(a) Since $P(x) = 0, Q(x) = x-2, R(x) = 0$，$a = 2$ is an ordinary point. Then we let the solution

$$y(x) = \sum_{n=0}^{\infty}c_{n}(x-2)^{n}$$

Thus,

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}(x-2)^{n-1}$$

$$y^{\prime\prime} = \sum_{n=2}^{\infty}n(n-1)c_{n}(x-2)^{n-2}$$

Put these into the given equation. Then we have

$$\sum_{n=2}^{\infty}n(n-1)c_{n}(x-2)^{n-2} + \sum_{n=0}^{\infty}c_{n}(x-2)^{n+1} = 0$$

Now align the power of $(x-2)$ to the least $n-2$. Then

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=3}^{\infty}c_{n-3}(x-2)^{n-2} = 0 .$$

Next align the series indexto the largest one. Then

$$2c_{2}(x-2) + \sum_{n=3}^{\infty}[n(n-1)c_{n} + c_{n-3}](x-2)^{n-2} = 0.$$

Using the term by term differentiation, we see that the coefficients of $(x-2)^{n}$ are all 0. Thus, we have the recurrence relation.

$$c_{2} = 0, n(n-1)c_{n} + c_{n-3} = 0 , n \geq 3$$

or

$$c_{2} = 0, c_{n} = -\frac{c_{n-3}}{n(n-1)}, n \geq 3$$

Note that $c_{0},c_{1}$ are determined by $y(0), y^{\prime}(0)$. Then we can think of these as arbitrary constants. Thus, we find $c_{2},c_{3},\ldots$ interms of $c_0, c_1$. Now

$$c_{2} = 0, c_{3} = \frac{-c_{1}}{3\cdot 2}, c_{4} = \frac{-c_{1}}{4\cdot 3} \cdots$$

Then

$$c_{3n} = \frac{c_{3n}}{c_{3n-3}}\frac{c_{3n-3}}{c_{3n-6}}\cdots \frac{c_{3}}{c_{0}}c_{0}$$

$$= \frac{(-1)}{3n(3n-1)}\cdot \frac{(-1)}{(3n-3)(3n-4)}\cdots \frac{(-1)}{3\cdot 2}c_{0}$$

$$= \frac{(-1)^n (3n-2)(3n-5)\cdots4 \cdot 1 c_{0}}{(3n)!}$$

or

$$c_{3n+1} = \frac{c_{3n+1}}{c_{3n-2}}\frac{c_{3n-2}}{c_{3n-5}}\cdots \frac{c_{4}}{c_{1}}c_{1}$$

$$= \frac{(-1)}{(3n+1)(3n)}\cdot \frac{(-1)}{(3n-2)(3n-3)}\cdots \frac{(-1)}{4\cdot 3}c_{0}$$

$$= \frac{(-1)^n (3n-1)(3n-4)\cdots 2 c_{1}}{(3n+1)!}$$

$$c_{3n+2} = \frac{c_{3n+2}}{c_{3n-1}}\frac{c_{3n-1}}{c_{3n-4}}\cdots \frac{c_{5}}{c_{2}}c_{2} = 0.$$

Thus, we have

$$y = \sum_{n=0}^{\infty}c_{3n}(x-2)^{3n} + \sum_{n=0}^{\infty}c_{3n+1}(x-2)^{3n+1} + \sum_{n=0}^{\infty}c_{3n+2}(x-2)^{3n+2}$$

$$= c_{0}\sum_{n=0}^{\infty} \frac{(-1)^n (3n-2)(3n-5)\cdots4 \cdot 1 }{(3n)!}(x-2)^{3n}$$

$$+ c_{1}\sum_{n=0}^{\infty} \frac{(-1)^n (3n-1)(3n-4)\cdots 2 }{(3n+1)!} (x-2)^{3n+1}$$

(b) Since $P(x) = x, Q(x) = 3, R(x) = x^2$，$a = 1$ is an ordinary point. Then let

$$y(x) = \sum_{n=0}^{\infty}c_{n}(x-1)^{n}$$

and

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}(x-1)^{n-1}$$

$$y^{\prime\prime} = \sum_{n=2}^{\infty}n(n-1)c_{n}(x-1)^{n-2}$$

Put these into the given equation. Then we have

$$\sum_{n=2}^{\infty}n(n-1)c_{n}(x-1)^{n-2} + \sum_{n=0}^{\infty}nc_{n}(x-1)^{n} + \sum_{n=0}^{\infty}3c_{n}(x-1)^{n} = x^2$$

Now align the power of $(x-1)$ to the least $n-2$. ，

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=3}^{\infty}(n-2)c_{n-2}(x-1)^{n-2} + \sum_{n=2}^{\infty}3c_{n-2}(x-1)^{n-2}$$

$$= (x-1)^2 + 2(x-1) + 1 .$$

Next align the series index to the $n=3$. Then

$$2c_{2} + 3c_{0} + \sum_{n=3}^{\infty}[n(n-1)c_{n} + (n-2)c_{n-2} + 3c_{n-2}](x-1)^{n-2} = (x-1)^2 + 2(x-1) + 1.$$

Both sides are equal. Thus, we have

$$2c_{2} + 3c_{0}= 1, 6c_{3} + c_{1} + 3c_{1} = 2, 12c_{4} + 2c_{2} + 3c_{2} = 1,$$

$$n(n-1)c_{n} + (n+1)c_{n-2} = 0 , n \geq 5$$

Note hat $c_{0},c_{1}$ are determined by $y(0), y^{\prime}(0)$. Then we can think of these as arbitray constants. Thus, we find $c_{2},c_{3},c_{4},\ldots$ interms of $c_-. c_1$. Then we have

$$c_{2} = \frac{1-3c_{0}}{2}, c_{3} = \frac{1-3c_{1}}{3}, c_{4} = \frac{-1 + 5c_{0}}{8}, c_{n} = -\frac{(n+1)c_{n-2}}{n(n-1)}, n \geq 5$$

$$c_{2n} = \frac{c_{2n}}{c_{2n-2}}\frac{c_{2n-2}}{c_{2n-4}}\cdots \frac{c_{6}}{c_{4}}c_{4}$$

$$= \frac{(-1)(2n+1)}{2n(2n-1)}\cdot \frac{(-1)(2n-1)}{(2n-2)(2n-3)}\cdots \frac{(-1)7}{6\cdot 5}c_{4}$$

$$= \frac{(-1)^n (2n+1)}{2^n n!}\frac{4\cdot 3}{-5}\cdot \frac{2\cdot 1}{-3}c_{4}$$

$$= \frac{(-1)^n (2n+1)}{2^n n!}\frac{4\cdot 3}{-5}\cdot \frac{2\cdot 1}{-3}(\frac{-1 + 5c_{0}}{8})$$

$$= \frac{(-1)^n (2n+1)}{2^n n!}(c_{0} - \frac{1}{5})$$

Also,

$$c_{2n+1} = \frac{c_{2n+1}}{c_{2n-1}}\frac{c_{2n-1}}{c_{2n-3}}\cdots \frac{c_{5}}{c_{3}}c_{3}$$

$$= \frac{(-1)(2n+2)}{(2n+1)(2n)}\cdot \frac{(-1)(2n)}{(2n-1)(2n-2)}\cdots \frac{(-1)6}{5 \cdot 4}c_{3}$$

$$= \frac{(-1)^n 2^n (n+1)!}{(2n+1)!}(\frac{3\cdot 2}{-4})c_{3}$$

$$= \frac{(-1)^n 2^n (n+1)!}{(2n+1)!}(\frac{3\cdot 2}{-4} \cdot \frac{2-4c_{1}}{6})$$

$$= \frac{(-1)^n 2^n (n+1)!}{(2n+1)!}(c_{1} - \frac{1}{2})$$

Thus, we have

$$y = \sum_{n=0}^{\infty}c_{2n}(x-1)^{2n} + \sum_{n=0}^{\infty}c_{2n+1}(x-1)^{2n+1}$$

$$= \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{2^n n!}(c_{0} - \frac{1}{5})(x-1)^{2n}$$

$$+ \sum_{n=0}^{\infty} \frac{(-1)^n 2^n (n+1)!}{(2n+1)!}(c_{1} - \frac{1}{2})(x-1)^{2n+1}$$