Problem 33 - Advanced Engineering Mathematics

Question

(a) \(\mathrm{By}\) inspection \(y=x\) and \(y=-x\) are solutions of the differential equation and not members of the family \(y=x \sin \left(\ln x+c_{2}\right).\) (b) Letting \(x=5\) and \(y=0\) in \(\sin ^{-1}(y / x)=\ln x+c_{2}\) we get \(\sin ^{-1} 0=\ln 5+c\) or \(c=-\ln 5 .\) Then \(\sin ^{-1}(y / x)=\ln x-\ln 5=\ln (x / 5) .\) Because the range of the arcsine function is \([-\pi / 2, \pi / 2]\) we must have $$\begin{array}{c} -\frac{\pi}{2} \leq \ln \frac{x}{5} \leq \frac{\pi}{2} \\ e^{-\pi / 2} \leq \frac{x}{5} \leq e^{\pi / 2} \\ 5 e^{-\pi / 2} \leq x \leq 5 e^{\pi / 2}. \end{array}$$ The interval of definition of the solution is approximately [1.04, 24.05].

Step-by-Step Solution

Answer

Interval is approximately [1.04, 24.05]; \(y = x\) and \(y = -x\) are not in the given family.

Step 1: Analyzing Given Solutions

We have to verify that both \(y = x\) and \(y = -x\) are solutions to some differential equation, although the problem doesn't specify which one right away. It's stated that these solutions don't belong to the family \(y=x \sin \left(\ln x+c_{2}\right)\). This means that substituting \(y = x\) or \(y = -x\) into the general family equation should not lead to a valid identity.

Step 2: Evaluating Specific Values

Substitute the given conditions, \(x = 5\) and \(y = 0\), into \(\sin^{-1}(y/x) = \ln x + c_2\). This yields \(\sin^{-1}(0) = \ln 5 + c_2\), resulting in \(c_2 = -\ln 5\) since \(\sin^{-1}(0) = 0\).

Step 3: Deriving from the Transformed Equation

We transform \(\sin^{-1}(y/x) = \ln x + c_2\) using the derived \(c_2\). Thus, \(\sin^{-1}(y/x) = \ln x - \ln 5 = \ln(x/5)\).

Step 4: Establish Range Constraints

We know the range of \(\sin^{-1}\) is \([-\pi/2, \pi/2]\), leading to \(-\pi/2 \leq \ln(x/5) \leq \pi/2\).

Step 5: Solve Inequality Bounds

Transforming the inequality \(-\pi/2 \leq \ln(x/5) \leq \pi/2\) into exponential terms, we get \(e^{-\pi/2} \leq x/5 \leq e^{\pi/2}\). Solving this gives \(5e^{-\pi/2} \leq x \leq 5e^{\pi/2}\).

Step 6: Approximate the Interval

Calculate the values \(5e^{-\pi/2} \approx 1.04\) and \(5e^{\pi/2} \approx 24.05\). Thus, the interval for \(x\) is approximately \([1.04, 24.05]\).

Key Concepts

Solution Verification

When verifying the solutions to any differential equation, the main task is to check if proposed solutions satisfy the equation. In our problem here, we are given two solutions: \(y = x\) and \(y = -x\). These are claimed to be solutions of the differential equation.

• To verify them, we would typically substitute each into the differential equation.
• If the original differential equation turns into a true statement after substitution, the given function is indeed a solution.
• This step ensures that our proposed solutions work with the given differential equation, providing confidence in proceeding with further calculations.

Interestingly, these solutions do not belong to the particular family described by \(y = x \sin(\ln x + c_2)\). This highlights that not every form of a solution fits into a family of solutions derived from a general differential equation.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\sin^{-1}\), play a crucial role in solving certain differential equations, especially when variables are intertwined with trigonometric expressions. Here, we deal with \(\sin^{-1}(y/x) = \ln x + c_2\).

• The inverse sine, \(\sin^{-1}\), tells us the angle whose sine equals a given value, constrained to the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• This range restriction is vital as it determines the valid interval for further computations.
• It's essential when dealing with real-world problems because it provides bounds for the solutions, ensuring mathematical viability and realism in the interpretations of the results.

Inequalities

Handling inequalities is all about managing the relationships between values to find ranges or sets that satisfy certain conditions. In this context, we transform \(-\pi/2 \leq \ln(x/5) \leq \pi/2\) into exponential terms.

• The natural logarithm inequality bounds lead to \(e^{-\pi/2} \leq x/5 \leq e^{\pi/2}\).
• This transformation is critical. Because we exponentiate both bounds, making sure we maintain the inequality, we move smoothly from a logarithmic to an exponential equation.
• The resulting range provides crucial insights into where the solution is valid, limiting the interval of definition.

Solving these inequalities is foundational when it comes to defining a solution's boundaries within a specific field of values.

Interval of Definition

The interval of definition is the set of values over which a particular solution to a differential equation is valid. For our problem, solving transformed inequalities enables us to approximate this interval.

• By solving \(5e^{-\pi/2} \leq x \leq 5e^{\pi/2}\), we determine that the interval is approximately \([1.04, 24.05]\).
• This means the solution holds true and is meaningful only within these endpoints.
• Establishing such intervals is crucial when applying mathematical models to real-world situations, ensuring that solutions remain realistic and applicable.

Therefore, understanding these boundaries helps predict where the solutions align with physical or theoretical contexts.