Verify $y=x^aZ_p\left(bx^c\right)$ is a solution to $y''+\left(\frac{1-2a}{x}\right)y'+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$

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?

Solution 1:

[2016-06-07] Note: General solution added to provide a comparison with the example part.

Here we show that according to OPs example \begin{align*} y(x)=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) \end{align*} is a solution of the differential equation \begin{align*} y^{\prime\prime}+9xy=0\tag{1} \end{align*}

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*}

We start with \begin{align*} y(x)&=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\\ \end{align*} and obtain the first derivative \begin{align*} y^\prime(x)&=\left(x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &=\left(x^{\frac{1}{2}}\right)^\prime Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^\frac{1}{2}\left(Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\tag{2}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^{\frac{1}{2}}{Z^\prime_{\frac{1}{3}}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\tag{3}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^{\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot3x^{\frac{1}{2}}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{4}\\ \end{align*}

Comment:

• In (2) we apply the product rule

• In (3) we apply the chain rule

The next step is getting the second derivative. We obtain from (4) \begin{align*} y^{\prime\prime}(x)&=\left(\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime +\left(3xZ^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &=\left(\frac{1}{2}x^{-\frac{1}{2}}\right)^\prime Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}\left(Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &\qquad+\left(3x\right)^\prime Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3x \left(Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\tag{5}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\\ &\qquad+3 Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\tag{6}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot 3x^{\frac{1}{2}}\\ &\qquad+3 Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot 3x^{\frac{1}{2}}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{7}\\ \end{align*}

Comment:

• In (5) we apply the product rule for both terms

• In (6) we apply the chain rule for both terms

So, let's put $y(x)=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})$ and the result (7) of $y^{\prime\prime}(x)$ into the differential equation (1). We obtain \begin{align*} y^{\prime\prime}&+9xy\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x\left(x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)\\ &=9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\left(9x^{\frac{3}{2}}-\frac{1}{4}x^{-\frac{3}{2}}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{8} \end{align*}

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*}

Substituting \begin{align*} x\rightarrow 2x^{\frac{3}{2}} \end{align*} in (9) results in \begin{align*} \left(2x^{\frac{3}{2}}\right)^2Z_{\frac{1}{3}}^{\prime\prime}(2x^{\frac{3}{2}})+2x^{\frac{3}{2}}Z_{\frac{1}{3}}^{\prime}(2x^{\frac{3}{2}})+\left((2x^{\frac{3}{2}})^2-\frac{1}{9}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0\\ 4x^3Z_{\frac{1}{3}}^{\prime\prime}(2x^{\frac{3}{2}})+2x^{\frac{3}{2}}Z_{\frac{1}{3}}^{\prime}(2x^{\frac{3}{2}})+\left(4x^3-\frac{1}{9}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0\\ \end{align*} and multiplying this equation with $\frac{9}{4}x^{-\frac{3}{2}}$ finally gives

\begin{align*}9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\left(9x^{\frac{3}{2}}-\frac{1}{4}x^{-\frac{3}{2}}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0 \end{align*}

We see the left-hand side is equal to (8) and we conclude that $x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(2x^{\frac{3}{2}}\right)$ is a solution of \begin{align*} y^{\prime\prime}+9xy=0 \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*}

Note: Here as small supplement to the already given nice answers by @JJacquelin and @Bacon the general derivation just to provide a comparison with the example part of my answer.

We show that
\begin{align*} y(x)=x^{a}Z_{p}(bx^{c}) \end{align*} is a solution of the differential equation \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\tag{10} \end{align*}

We start with calculating the first derivative \begin{align*} y^\prime(x)&=\left(x^{a}Z_{p}(bx^{c})\right)^\prime\\ &=\left(x^{a}\right)^\prime Z_{p}(bx^{c}) +x^a\left(Z_{p}(bx^{c})\right)^\prime\tag{11}\\ &=ax^{a-1}Z_{p}(bx^{c}) +x^{a}Z^\prime_p(bx^{c})\cdot\left(bx^{c}\right)^\prime\tag{12}\\ &=ax^{a-1}Z_{p}(bx^{c}) +x^{a}Z^\prime_p(bx^{c})\cdot bcx^{c-1}\\ &=ax^{a-1}Z_{p}(bx^{c}) +bcx^{a+c-1}Z^\prime_p(bx^{c})\tag{13}\\ \end{align*}

Comment:

• In (11) we apply the product rule

• In (12) we apply the chain rule

The next step is getting the second derivative. We obtain from (13) \begin{align*} y^{\prime\prime}(x)&=\left(ax^{a-1}Z_{p}(bx^{c})\right)^\prime +\left(bcx^{a+c-1}Z^\prime_{p}(bx^{c})\right)^\prime\\ &=(ax^{a-1})^\prime Z_{p}(bx^{c})+ax^{a-1}\left(Z_{p}(bx^{c})\right)^\prime\\ &\qquad+(bcx^{a+c-1})^\prime Z_{p}^\prime(bx^{c})+bcx^{a+c-1}\left(Z_{p}^\prime(bx^{c})\right)^\prime\tag{14}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+ax^{a-1}Z_{p}^\prime(bx^{c})\cdot\left(bx^{c}\right)^\prime\\ &\qquad+bc(a+c-1)x^{a+c-2} Z_{p}^\prime(bx^{c}) +bcx^{a+c-1}Z_{p}^{\prime\prime}(bx^{c})\cdot\left(bx^{c}\right)^\prime\tag{15}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+ax^{a-1}Z_{p}^\prime(bx^{c})\cdot bcx^{c-1}\\ &\qquad+bc(a+c-1)x^{a+c-2} Z_{p}^\prime(bx^{c}) +bcx^{a+c-1}Z_{p}^{\prime\prime}(bx^{c})\cdot bcx^{c-1}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+(2a+c-1)bcx^{a+c-{2}}Z_{p}^\prime(bx^{c})\\ &\qquad+b^2c^2x^{a+2c-2}Z_{p}^{\prime\prime}(bx^{c})\tag{16}\\ \end{align*}

Comment:

• In (14) we apply the product rule for both terms

• In (15) we apply the chain rule for both terms

So, let's put $y(x)=x^{a}Z_{p}(bx^{c})$ and (13), the result of $y^\prime(x)$ and (16), the result of $y^{\prime\prime}(x)$ into the differential equation (10). We obtain \begin{align*} y^{\prime\prime}(x)&+\left(\frac{1-2a}{x}\right)y^\prime(x)+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y(x)\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+(2a+c-1)bcx^{a+c-2}Z_{p}^\prime(bx^{c})\\ &\qquad+b^2c^2x^{a+2c-2}Z_{p}^{\prime\prime}(bx^{c})\\ &\qquad+\left(\frac{1-2a}{x}\right)\left(ax^{a-1}Z_{p}(bx^{c}) +bcx^{a+c-1}Z^\prime_p(bx^{c})\right)\\ &\qquad+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]x^{a}Z_{p}(bx^{c})\\ &=c^2x^{a-2}\left(b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^cZ_{p}^{\prime}(bx^{c}) +(b^2x^{2c}-p^2)Z_{p}(bx^{c})\right)\tag{17} \end{align*}

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*}

Substituting \begin{align*} x\rightarrow bx^{c} \end{align*} in (18) results in \begin{align*} (bx^{c})^2Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left((bx^{c})^2-p^2\right)Z_{p}(bx^{c})=0\\ b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left(b^2x^{2c}-p^2\right)Z_{p}(bx^{c})=0 \end{align*} and multiplying this equation with $c^2x^{a-2}$ finally gives

\begin{align*}c^2x^{a-2}\left(b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left(b^2x^{2c}-p^2\right)Z_{p}(bx^{c})\right)=0 \end{align*}

We see the left-hand side is equal to (17) and we conclude that $x^{a}Z_{p}\left(bx^{c}\right)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*}

Solution 2:

This answer is too long for a comment and refers to BLAZE's comment above as to how to compute $$\frac{d}{dx}\frac{dY}{dX}$$

Setting the Scene $y(x)=x^{\alpha}Y(X),\quad X=\beta x^{\gamma}$ Now it is obvious that \begin{align} \frac{dy}{dx} &= \alpha x^{\alpha - 1}Y+x^{\alpha}\frac{dY}{dx} \\ &= \alpha x^{\alpha-1}Y+x^{\alpha}\frac{dY}{dX}\frac{dX}{dx} \end{align} We could expand as JJacquelin does but let's leave the calculation here. Now computing the second derivative \begin{align} \frac{d^{2}y}{dx^{2}} &= \frac{d}{dx}\left(\alpha x^{\alpha-1}Y+x^{\alpha}\frac{dY}{dX}\frac{dX}{dx} \right) \\ &= \alpha (\alpha - 1)x^{\alpha-1}Y+\alpha x^{\alpha-1}\frac{dY}{dx}+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx} + x^{\alpha}\frac{d}{dx}\left(\frac{dY}{dX}\frac{dX}{dx}\right) \\ &= \alpha (\alpha - 1)x^{\alpha-1}Y+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+x^{\alpha}\frac{d^{2}Y}{dX^{2}}\left(\frac{dX}{dx}\right)^{2}+x^{\alpha}\frac{dY}{dX}\frac{d^{2}X}{dx^{2}}\\ &=\alpha (\alpha - 1)x^{\alpha-1}Y+2\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+x^{\alpha}\frac{d^{2}Y}{dX^{2}}\left(\frac{dX}{dx}\right)^{2}+x^{\alpha}\frac{dY}{dX}\frac{d^{2}X}{dx^{2}} \end{align} I hope this clears things up a little for BLAZE.

Apologies to the moderators with regards to this not being a full answer to the question, but I thought the level of computation was too long for a comment.