Chapter 5, Problem 1CR

To determine

Whether the given statement is true or false.

Answer to Problem 1CR

The given statement is $\underset{\_}{\text{False}}$.

Explanation of Solution

Calculation:

Consider the Bessel’s equation of order 1,

$x^2{y}^{″}+xy\text{'}+\left(x^2-1\right)y=0$        (1)

Since $x=0$ is regular singular point of Bessel’s equation

So, there exists at least one solution of the form $y={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+r}}$.

Substitute the equation (1) in above expression,

$\begin{array}{c}{x}^2{y}^{″}+xy\text{'}+\left(x^2-1\right)y=\left(\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{c}_n\left(n+r\right)\left(n+r-1\right)x^{n+r}}+{\displaystyle \sum _{n=0}^{\infty }{c}_n\left(n+r\right)x^{n+r}}+\\ {\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+r+2}}-{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+r}}\end{array}\right)\\ =\left(\begin{array}{l}{c}_0\left(r^2-r+r-1\right)x^r+x^r{\displaystyle \sum _{n=0}^{\infty }{c}_n\left[\left(n+r\right)\left(n+r-1\right)+\left(n+r\right)-1\right]}{x}^n+\\ x^r{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+2}}\end{array}\right)\\ =c_0\left(r^2-1\right)x^r+x^r{\displaystyle \sum _{n=0}^{\infty }{c}_n\left[{\left(n+r\right)}^2-1\right]}{x}^n+x^r{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+2}}\end{array}$

From above the identical equation is $r^2-1=0$.

The identical roots are $r_1=1,r_2=1$.

So, $J_1\left(x\right)\text{ and }{J}_{-1}\left(x\right)$ are not linearly independent when roots are positive integer

Thus, the general solution of $x^2{y}^{″}+xy\text{'}+\left(x^2-1\right)y=0$ is $y=c_1{J}_1\left(x\right)+c_2{Y}_1\left(x\right)$

Therefore, the given statement is $\underset{\_}{\text{False}}$.