Problem 17 - Advanced Engineering Mathematics

Question

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ y^{\prime}=2 x y^{2} ; \quad y=1 /\left(4-x^{2}\right) $$

Step-by-Step Solution

Answer

The function \( y = \frac{1}{4-x^2} \) is an explicit solution to the differential equation \( y' = 2xy^2 \) on the interval where \( x \neq \pm 2 \).

Step 1: Differentiate the given function

To verify that the given function is a solution, we first need to find its derivative. The given function is \( y = \frac{1}{4-x^2} \). Use the quotient rule to differentiate: \[y' = \frac{d}{dx}\left(\frac{1}{4-x^2}\right) = \frac{0(4-x^2) - 1(0 - 2x)}{(4-x^2)^2} = \frac{2x}{(4-x^2)^2}\]

Step 2: Check the initial differential equation

The given differential equation is \( y' = 2xy^2 \). We have already found \( y' \) to be \( \frac{2x}{(4-x^2)^2} \) in Step 1. Now, substitute \( y = \frac{1}{4-x^2} \) into the right-hand side of the differential equation to compare both sides.

Step 3: Substitute and compare

Substitute \( y = \frac{1}{4-x^2} \) into the right-hand side of the differential equation. \[2x \left(\frac{1}{4-x^2}\right)^2 = 2x \frac{1}{(4-x^2)^2} = \frac{2x}{(4-x^2)^2}\]Compare this with the derivative found in Step 1, which is \( \frac{2x}{(4-x^2)^2} \). Both expressions are identical.

Step 4: Determine the domain

The function \( y \) is defined as \( y = \frac{1}{4-x^2} \), which is undefined when \( 4-x^2=0 \) or \( x = \pm 2 \). Therefore, the appropriate interval \( I \) is where \( x eq \pm 2 \).

Key Concepts

Explicit Solution

In the realm of differential equations, an explicit solution is a function that directly expresses the dependent variable as a function of the independent variable.
In simpler terms, if you have a differential equation involving a function and its derivatives, an explicit solution will look like "y = some function of x." For example, consider the exercise provided. The function given is \( y = \frac{1}{4-x^2} \). This is an explicit formulation where 'y' is expressed in terms of 'x'. Another point of interest is that explicit solutions make it straightforward to plug in values for 'x' and calculate corresponding values for 'y'. This helps in visualizing the behavior of the function across the specified domain. When dealing with real-world problems, explicit solutions make it easier to analyze and apply the results.

Quotient Rule

The quotient rule is a technique in calculus used to differentiate functions that are expressed as a quotient, or fraction, of two other functions. When you have a function \( f(x) \) divided by another function \( g(x) \), the quotient rule helps you find the derivative efficiently. The quotient rule states:

• Given \( y = \frac{u(x)}{v(x)} \), the derivative \( y' \) is:

\[y' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\]In the example from the exercise, the function \( y = \frac{1}{4-x^2} \) was differentiated using this rule. You see each part being used: the numerator becomes zero (as \( u = 1 \) is constant and \( u' = 0 \)), and for the denominator, it is squared. The result wraps up the task by giving us the derivative \( y' = \frac{2x}{(4-x^2)^2} \). This derived function \( y' \) can then be verified against the differential equation provided, demonstrating the utility of the quotient rule in solving such problems.

Domain of Function

Finding the domain of a function involves identifying all possible values of 'x' for which the function is defined. It is crucial to check any restrictions that may arise from operations such as division by zero, square roots of negative numbers, etc. For instance, in the given function \( y = \frac{1}{4-x^2} \), it includes a denominator \((4 - x^2)\). If the denominator equals zero, the function becomes undefined. Thus,

• we set \( 4 - x^2 = 0 \)
• solve to find \( x = \pm 2 \)

These are the points where the function "breaks," so to speak. Hence, the domain of this function is all real numbers except \( x = 2 \) and \( x = -2 \). Understanding the domain is essential because it tells you where the solution holds and helps avoid misinterpretation of results, especially when applying these findings in context-driven scenarios.