Problem 8 Change the independent variable ... [FREE SOLUTION]

Chapter 4: Problem 8

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

Short Answer

Expert verified

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

Step by step solution

Define the New Variable

Let鈥檚 start by defining the new independent variable. Here, it鈥檚 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: \[\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{2}{u} \cdot \frac{dy}{du} \right).\]Use $ \frac{d}{dx} = \frac{d}{du} \cdot \frac{du}{dx} = \frac{d}{du} \cdot \frac{2}{u} $. So, \[\frac{d^2 y}{dx^2} = \frac{d}{du} \left( \frac{2}{u} \cdot \frac{dy}{du} \right) \cdot \frac{2}{u}\].

Simplify the Second Derivative

Simplify further: \[\frac{d^2 y}{dx^2} = \frac{2}{u} \cdot \frac{d}{du}\left( \frac{2}{u} \cdot \frac{dy}{du} \right) = \frac{2}{u} \cdot \left( -\frac{2}{u^2} \cdot \frac{dy}{du} + \frac{2}{u} \cdot \frac{d^2 y}{du^2} \right)\]\[= -\frac{4}{u^3} \cdot \frac{dy}{du} + \frac{4}{u^2} \cdot \frac{d^2 y}{du^2}.\]

Substitute Into the Original Equation

Now, substitute these into the original Bessel equation: \[x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - (1 - x) y = 0\]. 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鈥檚 put all together in the equation: \[\frac{u^4}{16} \cdot \left( -\frac{4}{u^3} \frac{dy}{du} + \frac{4}{u^2} \cdot \frac{d^2 y}{du^2} \right) + \frac{u^2}{4} \cdot \frac{2}{u} \cdot \frac{dy}{du} - \left( 1 - \frac{u^2}{4} \right)y = 0.\]Simplifying: \[-\frac{u}{4} \cdot \frac{dy}{du} + \frac{u^2}{4} \frac{d^2 y}{du^2} + \frac{u}{2} \cdot \frac{dy}{du} - y + \frac{u^2}{4}y = 0.\]

Combine Like Terms

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

Multiply Through by 4

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:\[x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}-(1-x) y=0\]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 $\frac{d^{2} y}{dx^{2}} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{du} \left( \frac{dy}{du} \cdot \frac{du}{dx} \right) \cdot \frac{du}{dx}$.Derivatives transform by using the chain rule, thereby simplifying multi-variable relationships in differential equations.

91影视

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