In order for the question that I have to make any sense I must first include some background information as given in my textbook:
The standard form of Bessel's differential equation is $$x^2y^{\prime\prime}+xy^{\prime} + (x^2 – p^2)y=0\tag{1}$$ where $(1)$ has a first solution given by $$\fbox{$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}$}\tag{2}$$ and a second solution given by $$\fbox{$J_{-p}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}$}\tag{3}$$ where $J_p(x)$ is called the Bessel function of the first kind of order $p$.
Although $J_{−p}(x)$ is a satisfactory second solution when $p$ is not an integer, it is customary to use a linear combination of $J_p(x)$ and $J_{−p}(x)$ as the second solution. Any combination of $J_p(x)$ and $J_{−p}(x)$ is a satisfactory second solution of Bessel’s equation. The combination which is used is called the Neumann (or the Weber) function and is denoted by $N_p(x)$ where $$N_p(x)=\frac{\cos(\pi p)J_p(x)-J_{-p}(x)}{\sin(\pi p)}\tag{4}$$
Full details on the derivation of $(2)$ as a solution to $(1)$ can be found here in my previous question.
Many differential equations occur in practice that are not of the standard form $(1)$ but whose solutions can be written in terms of Bessel functions. It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{5}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{6}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are
constants.To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{7}$$ If $(7)$ is of the type $(5)$, then we must have $$1-2a=0$$ $$2(c-1)=1$$ $$(bc)^2=9$$ $$a^2-p^2c^2=0$$ from these $4$ equations we find
$$a=\dfrac12$$ $$c=\dfrac32$$ $$b=2$$ $$p=\dfrac{a}{c}=\dfrac13$$Then the solution of $(7)$ is $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)\tag{8}$$ This means that the general solution of $(7)$ is $$y=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]\tag{9}$$ where $A$ and $B$ are arbitrary constants.
Finally,$\color{#180}{\text{ my goal is to show that }}$${(6)}$ $\color{#180}{\text{is a solution to }}$$(5)$.
However to gain some insight, I must first be able to show that $(8)$ or $(9)$ is a solution to $(7)$.
So my attempt goes as follows:
So I need to compute $y^{\prime\prime}$ or at least to begin with, $y^{\prime}$; It is at this point where I am immediately stuck as I do not understand how to differentiate $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)\tag{8}$$ as I'm confused as to how to take the derivative of the $Z_{1/3}\left(2x^{3/2}\right)$ factor.
Could someone please provide some hints or advice on how I would go about carrying out this differentiation?
Best Answer
[2016-06-07] Note: General solution added to provide a comparison with the example part.
In the following we use the prime notation $f^\prime$ to denote the derivative of $f$. We will often use the product rule for derivatives \begin{align*} (f\cdot g)^\prime=f^\prime \cdot g+f\cdot g^\prime\qquad\qquad (f(x)g(x))^\prime=f^\prime(x) g(x)+f(x)g^\prime(x) \end{align*} and the chain rule \begin{align*} (f\circ g)^\prime=(f^\prime \circ g)\cdot g^\prime\qquad\qquad\quad (f(g(x)))^\prime=f^\prime(g(x))g^\prime(x) \end{align*}
Comment:
In (2) we apply the product rule
In (3) we apply the chain rule
Comment:
In (5) we apply the product rule for both terms
In (6) we apply the chain rule for both terms
On the other hand we know that $Z_p(x)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*} This means $Z_{\frac{1}{3}}(x)$ fulfils \begin{align*} x^2Z_{\frac{1}{3}}^{\prime\prime}(x)+xZ_{\frac{1}{3}}^{\prime}(x)+\left(x^2-\frac{1}{9}\right)Z_{\frac{1}{3}}(x)=0\tag{9} \end{align*}
Note: In the same way we can show that \begin{align*} y(x)=x^aZ_p(bx^c) \end{align*} is a solution of \begin{align*} y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^\prime+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y=0 \end{align*}
Comment:
In (11) we apply the product rule
In (12) we apply the chain rule
Comment:
In (14) we apply the product rule for both terms
In (15) we apply the chain rule for both terms
On the other hand we know that $Z_p(x)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*} This means $Z_{p}(x)$ fulfils \begin{align*} x^2Z_{p}^{\prime\prime}(x)+xZ_{p}^{\prime}(x)+\left(x^2-p^2\right)Z_{p}(x)=0\tag{18} \end{align*}