Chapter 17.5, Problem 1E

To determine

To find: The general solution of the differential equation $y^{″}+2y^{\prime }=0$ by using power series.

Answer to Problem 1E

The general solution of the differential equation $y^{″}+2y^{\prime }=0$ is $\underset{\_}{y\left(x\right)=a+b{e}^{-2x}}$.

Explanation of Solution

Formula used:

Power Series Method:

The power series method for solving a second-order homogenous differential equation consists of finding the coefficients of a power series $y\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n=c_0+c_{1}x+c_2{x}^2+\dots }$

Calculation:

The given equation is $y^{″}+2y^{\prime }=0$.

Assume that the series solution takes the form of $y={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}$.

Calculate the derivative of $y={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}$.

$y^{\prime }={\displaystyle \sum _{n=1}^{\infty }n{c}_n{x}^{n-1}}$

Now calculate the second derivative of $y={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}$.

$\begin{array}{c}{y}^{″}={\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)c_n{x}^{n-1-1}}\\ ={\displaystyle \sum _{n=2}^{\infty }n\left(n-1\right)c_n{x}^{n-2}}\end{array}$

Substitute the derivatives of $y^{″}$ and $y^{\prime }$ in $y^{″}+2y^{\prime }=0$ as follows.

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

Equate the coefficient of each power of x to zero as shown in the following table.

Power of x	Coefficient equation
$x^0$	$2\left(1\right)c_2+2c_1=0$ or $c_2=-c_1$
$x^1$	$3\left(2\right)c_3+2\left(2\right)c_2=0$ or $c_3=-\frac{2}{3}{c}_2=\frac{2}{3}{c}_1$
$x^2$	$4\left(3\right)c_4+2\left(3\right)c_3=0$ or $c_4=-\frac{1}{2}{c}_3=-\frac{1}{3}{c}_1$
$x^3$	$5\left(4\right)c_5+2\left(4\right)c_4=0$ or $c_5=-\frac{2}{5}{c}_4=\frac{2}{15}{c}_1$
$x^4$	$6\left(5\right)c_6+2\left(5\right)c_5=0$ or $c_6=-\frac{1}{3}{c}_5=-\frac{2}{45}{c}_1$
$x^n$	$\left(n+2\right)\left(n+1\right)c_{n+2}+2\left(n+1\right)c_{n+1}=0$ or $c_{n+2}=-\frac{2}{n+2}{c}_{n+1}$

The recursive relation $c_{n+2}=-\frac{2}{n+2}{c}_{n+1}$ can be written as $c_n=-\frac{2}{n}{c}_{n-1}$.

Thus,

$\begin{array}{c}{c}_n=\left(-\frac{2}{n}\right)\left(-\frac{2}{n-1}{c}_{n-2}\right)\\ =\left(-\frac{2}{n}\right)\left(-\frac{2}{n-1}\right)\left(-\frac{2}{n-2}{c}_{n-3}\right)\\ =\frac{{\left(-2\right)}^{n-1}}{n!}{c}_1\left(n\ge 2\right)\end{array}$

Hence,

$\begin{array}{c}y=c_0+c_{1}x-c_1{x}^2+\frac{2}{3}{c}_1{x}^3-\frac{1}{3}{c}_1{x}^4+\frac{2}{15}{c}_1{x}^5-\frac{2}{45}{c}_1{x}^6+\cdots \\ =c_0+\frac{c_1}{2}-\frac{c_1}{2}+\frac{c_1\left(2x\right)}{2}-\frac{c_1}{2}\frac{{\left(2x\right)}^2}{2!}+\frac{c_1}{2}\frac{{\left(2x\right)}^3}{3!}-\frac{c_1}{2}\frac{{\left(2x\right)}^4}{4!}+\frac{c_1}{2}\frac{{\left(2x\right)}^5}{5!}-\frac{c_1}{2}\frac{{\left(2x\right)}^6}{6!}+\cdots \\ =\left(c_0+\frac{c_1}{2}\right)-\frac{c_1}{2}\left(1-\left(2x\right)+\frac{{\left(2x\right)}^2}{2!}-\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^4}{4!}-\frac{{\left(2x\right)}^5}{5!}+\frac{{\left(2x\right)}^6}{6!}-\cdots \right)\\ =c_0+\frac{c_1}{2}-\frac{c_1}{2}{\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-2x\right)}^n}{n!}}\end{array}$

On further simplification,

$\begin{array}{c}y=\left(c_0+\frac{c_1}{2}\right)-\frac{c_1}{2}{e}^{-2x}\\ =a+b{e}^{-2x}\end{array}$

Where $a=c_0+\frac{c_1}{2}$ and $b=-\frac{c_1}{2}$.

Therefore, the general solution of the differential equation $y^{″}+2y^{\prime }=0$ is $\underset{\_}{y\left(x\right)=a+b{e}^{-2x}}$.