Question

33. (a) The differential equation ${\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{\lambda y}\mathbf{=}\mathbf{0}$has an irregular singular point at x = 0 Show that the substitution t = 1/x yields the DE.

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{t}}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{\lambda y}\mathbf{=}\mathbf{0}$ which now has a regular singular point at t =0

(b) Use the method of this section to find two series solutions of the second equation in part (a) about the regular singular point t = 0

(c) Express each series solution of the original equation in terms of elementary functions.

Short answer

The answers are:

1. a ) It has a regular singular point at
2. b) $\mathrm{y}={\mathrm{C}}_1\sum _{\mathrm{n}=0}^{\infty }\frac{-{\mathrm{\lambda }}^{\mathrm{n}}{\mathrm{c}}_0}{(2\mathrm{n}+1)!}{\mathrm{t}}^{2\mathrm{n}}+{\mathrm{C}}_2{\mathrm{t}}^{-1}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-\mathrm{\lambda })}^{\mathrm{n}}{\mathrm{c}}_0}{\left(2\mathrm{n}\right)!}{\mathrm{t}}^{2\mathrm{n}}$
   
3. c)$\mathrm{y}={\mathrm{C}}_1\sum _{\mathrm{n}=0}^{\infty }\frac{-{\mathrm{\lambda }}^{\mathrm{n}}{\mathrm{c}}_0}{(2\mathrm{n}+1)!}{\left(\frac{1}{\mathrm{x}}\right)}^{2\mathrm{n}}+{\mathrm{C}}_2\mathrm{x}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-\mathrm{\lambda })}^{\mathrm{n}}{\mathrm{c}}_0}{\left(2\mathrm{n}\right)!}(\frac{1}{\mathrm{x}}{)}^{2\mathrm{n}}$
   

Step-by-step solution

Step 1: Part a)

We have $x^4\frac{d^{2}y}{d{x}^2}+\lambda y=0$

Which has an irregular singular point at x = 0. Our aim is to use the substitution into t = 1/x the given differential equation and find to what it leads. First, we have a composite function. Therefore, we need to use the chain rule as follows;

localid="1664865136421" $\begin{array}{rcl}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}& =& \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{dy}}{\mathrm{dt}}\frac{\mathrm{dt}}{\mathrm{dx}}\right]\\ & =& \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{dy}}{\mathrm{dt}}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{dt}}{\mathrm{dx}}\right]\\ & =& \left[\frac{\mathrm{dt}}{\mathrm{dx}}\frac{\mathrm{d}}{\mathrm{dt}}\right]\left[\frac{\mathrm{dy}}{\mathrm{dt}}\right]\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{dy}}{\mathrm{dt}}\frac{{\mathrm{d}}^2\mathrm{t}}{{\mathrm{dx}}^2}\\ & =& \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}{\left[\frac{\mathrm{dt}}{\mathrm{dx}}\right]}^2+\frac{\mathrm{dy}}{\mathrm{dt}}\frac{{\mathrm{d}}^2\mathrm{t}}{{\mathrm{dx}}^2}...\left(1\right)\end{array}$

We find the first and second derivative of t(x)

$$\begin{gathered} \frac{dt}{dx}=\frac{-1}{x^2}...\left(2\right) \\ \frac{d^{2}t}{d{x}^2}=\frac{2}{x^3}...\left(3\right) \end{gathered}$$

Substitute the (2) and (3) into (1), yields

localid="1664316815643" $$\begin{gathered} \frac{dt}{dx}=\frac{-1}{x^2}...\left(2\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{t}^2}[\frac{-1}{x^2}{]}^2+\frac{2}{x^3}\frac{dy}{dt} \\ =\frac{1}{x^4}\frac{d^{2}y}{d{t}^2}+\frac{2}{x^3}\frac{dy}{dt}...\left(4\right) \end{gathered}$$

We substitute (4) into the given equation, yields

localid="1664865166620" $\begin{array}{rcl}{\mathrm{x}}^4\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{\lambda y}& =& 0\\ {\mathrm{x}}^4\left[\frac{1}{{\mathrm{x}}^4}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{2}{{\mathrm{x}}^3}\frac{\mathrm{dy}}{\mathrm{dt}}\right]+\mathrm{\lambda y}& =& 0\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{2}{\mathrm{t}}\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{\lambda y}& =& 0...\left(5\right)\end{array}$

Which has a regular singular point at t =0.

 Part b)

We state, without a proof, that if we have a differential equation, with the following standard form;

$y\text{'}\text{'}+P\left(t\right)y\text{'}+Q\left(t\right)=0$

Which has a regular singular point at t = t₀, then we can use Forbenious Method to solve this differential equation. We assume the solution, according to Forbenious, to be as follows;

$y=\sum _{n=0}^{\infty }{c}_n{t}^{n+r}$

We find the first and second derivative.

$$\begin{gathered} y\text{'}=\sum _{n=0}^{\infty }{c}_n(n+r)t^{n+r-1} \\ y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)t^{n+r-2} \end{gathered}$$

Therefore, the solution for r₂ = -1 becomes as follows

$$\begin{gathered} y_2=t^{-1}\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}{t}^{2n} \\ =c_0{t}^{-1}\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n}{\left(2n\right)!}{t}^{2n} \end{gathered}$$

We can obtain a general solution by using Superposition Principle as follows

$y=C_1\sum _{n=0}^{\infty }\frac{-{\lambda }^n{c}_0}{(2n+1)!}{t}^{2n}+C_2{t}^{-1}\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}{t}^{2n}$

Step 3: Part c):

To express the series solutions, we obtained in part b) in terms of the elementary function, we substitute t = 1/x into our solution.

$$\begin{gathered} y=C_1\sum _{n=0}^{\infty }\frac{-{\lambda }^n{c}_0}{(2n+1)!}{t}^{2n}+C_2{t}^{-1}\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}{t}^{2n} \\ =C_1\sum _{n=0}^{\infty }\frac{-{\lambda }^n{c}_0}{(2n+1)!}{\left(\frac{1}{x}\right)}^{2n}+C_{2}x\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}{\left(\frac{1}{x}\right)}^{2n} \end{gathered}$$

Therefore, finally we get

1. a) It has a regular singular point at
2. b)$y=C_1\sum _{n=0}^{\infty }\frac{-{\lambda }^n{c}_0}{(2n+1)!}{t}^{2n}+C_2{t}^{-1}\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}{t}^{2n}$
3. c)$y=C_1\sum _{n=0}^{\infty }\frac{-{\lambda }^n{c}_0}{(2n+1)!}{\left(\frac{1}{x}\right)}^{2n}+C_{2}x\sum _{n=0}^{\infty }\frac{{(-\lambda )}^n{c}_0}{\left(2n\right)!}(\frac{1}{x}{)}^{2n}$