Question

Change the independent variable from $x$ to $u=2\sqrt{x}$ in the Bessel equation $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$ and show that the equation becomes $$u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0.$$

Short answer

The equation becomes $u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0$.

Step-by-step solution

Define the New Variable

Let’s start by defining the new independent variable. Here, it’s given that $u=2\sqrt{x}$.

Express x in Terms of u

To proceed, we need to express $x$ in terms of $u$. Since $u=2\sqrt{x}$, squaring both sides, we get $u^2=4x$. Solving for $x$, we have $x=\frac{u^2}{4}$.

Compute the Derivatives

To transform the equation, we need the first and second derivatives of $y$ with respect to $u$. First, find $\frac{dx}{du}$: $$\frac{dx}{du}=\frac{d}{du}\left(\frac{u^2}{4}\right)=\frac{u}{2}$$.

Chain Rule Application for First Derivative

Use the chain rule to find $\frac{dy}{dx}$. $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$. Substitute $\frac{du}{dx}=\frac{1}{\frac{dx}{du}}=\frac{2}{u}$, so $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{2}{u}$$.

Chain Rule for Second Derivative

For the second derivative, apply the product rule and chain rule: $$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\frac{2}{u}\cdot \frac{dy}{du}).$$Use $\frac{d}{dx}=\frac{d}{du}\cdot \frac{du}{dx}=\frac{d}{du}\cdot \frac{2}{u}$. So, $$\frac{d^{2}y}{d{x}^2}=\frac{d}{du}(\frac{2}{u}\cdot \frac{dy}{du})\cdot \frac{2}{u}$$.

Simplify the Second Derivative

Simplify further: $$\frac{d^{2}y}{d{x}^2}=\frac{2}{u}\cdot \frac{d}{du}(\frac{2}{u}\cdot \frac{dy}{du})=\frac{2}{u}\cdot (-\frac{2}{u^2}\cdot \frac{dy}{du}+\frac{2}{u}\cdot \frac{d^{2}y}{d{u}^2})$$$$=-\frac{4}{u^3}\cdot \frac{dy}{du}+\frac{4}{u^2}\cdot \frac{d^{2}y}{d{u}^2}.$$

Substitute Into the Original Equation

Now, substitute these into the original Bessel equation: $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$. Since $x=\frac{u^2}{4}$, we know that $x^2={\left(\frac{u^2}{4}\right)}^2=\frac{u^4}{16}$.

Simplify and Rearrange

Let’s put all together in the equation: $$\frac{u^4}{16}\cdot (-\frac{4}{u^3}\frac{dy}{du}+\frac{4}{u^2}\cdot \frac{d^{2}y}{d{u}^2})+\frac{u^2}{4}\cdot \frac{2}{u}\cdot \frac{dy}{du}-(1-\frac{u^2}{4})y=0.$$Simplifying: $$-\frac{u}{4}\cdot \frac{dy}{du}+\frac{u^2}{4}\frac{d^{2}y}{d{u}^2}+\frac{u}{2}\cdot \frac{dy}{du}-y+\frac{u^2}{4}y=0.$$

Combine Like Terms

Combine the terms with $\frac{dy}{du}$: $$\frac{u^2}{4}\frac{d^{2}y}{d{u}^2}+(-\frac{u}{4}+\frac{u}{2})\frac{dy}{du}+(-1+\frac{u^2}{4})y=0.$$Simplify further: $$\frac{u^2}{4}\frac{d^{2}y}{d{u}^2}+\frac{u}{4}\frac{dy}{du}+(\frac{u^2}{4}-1)y=0.$$

Multiply Through by 4

To clear the fractions, multiply the entire equation by 4: $$u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0.$$This is the desired equation.

Key concepts

Differential Equations

Differential equations are equations that relate a function to its derivatives. They are crucial in describing various natural phenomena, such as population growth, heat transfer, and wave propagation. The Bessel equation, a specific type of differential equation, is often used in problems requiring cylindrical symmetry. A typical Bessel equation looks like this:$$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$To solve differential equations, various techniques and transformations can be applied. One common method is a change of variables when the equation is too complex to solve in its original form.

Change of Variables

A change of variables introduces a new variable to simplify an equation. In this context, we change the independent variable from $x$ to $u=2\sqrt{x}$. This transformation helps simplify the equation by leveraging the relationships between the derivatives of the original and new variables. **Steps to Follow: **1. **Define the New Variable:** Here, $u=2\sqrt{x}$.2. **Express Old Variable in Terms of New Variable:** Since $u=2\sqrt{x}$, we get $x=\frac{u^2}{4}$ by squaring both sides and solving for $x$.3. **Compute the Derivatives:** Use the chain rule to find how derivatives of $y$ with respect to $x$ relate to those with respect to $u$.This method converts a complex differential equation into one that may be more easily solvable, often revealing patterns or symmetries that were not initially evident.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When you need to change variables in a differential equation, the chain rule helps connect the derivatives with respect to the old variable to those with respect to the new variable. **Steps to Use Chain Rule: **1. **First Derivative:** For $y(x)$, we have $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$. Given the relationship $u=2\sqrt{x}$, finding $\frac{dx}{du}$ allows us to transform this derivative.2. **Second Derivative:** Apply the chain rule and product rule together to find the second derivative. For $y(x)$, we have $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{du}(\frac{dy}{du}\cdot \frac{du}{dx})\cdot \frac{du}{dx}$.Derivatives transform by using the chain rule, thereby simplifying multi-variable relationships in differential equations.