Problem 45 - Advanced Engineering Mathematics

Question

We are looking for a function \(y(x)\) such that \\[y^{2}+\left(\frac{d y}{d x}\right)^{2}=1\\]. Using the positive square root gives \\[\frac{d y}{d x}=\sqrt{1-y^{2}} \Longrightarrow \frac{d y}{\sqrt{1-y^{2}}}=d x \Longrightarrow \sin ^{-1} y=x+c\\]. Note that when \(c=c_{1}=0\) and when \(c=c_{1}=\pi / 2\) we obtain the well known particular solutions \(y=\sin x\) \(y=-\sin x, y=\cos x,\) and \(y=-\cos x .\) Note also that \(y=1\) and \(y=-1\) are singular solutions.

Step-by-Step Solution

Answer

The solutions are \(y = \sin x, y = -\sin x, y = \cos x, y = -\cos x, y = 1, y = -1\).

Step 1: Understand the Given Equation

We start with the differential equation \(y^{2} + \left( \frac{dy}{dx} \right)^{2} = 1\). This equation describes a relationship between the function \(y\) and its derivative \(\frac{dy}{dx}\). Our task is to find function \(y(x)\) such that the equation holds.

Step 2: Differentiate with respect to x

Recognize that we can explore the form \(\frac{dy}{dx} = \sqrt{1-y^2} \) based on the constraint \(y^{2} + \left( \frac{dy}{dx} \right)^{2} = 1\). Notice that this equation resembles the Pythagorean identity and considers the range for a trigonometric function.

Step 3: Separate the Variables

Recast the equation \(\frac{dy}{dx} = \sqrt{1-y^2}\) into a separable form: \(\frac{dy}{\sqrt{1-y^2}} = dx\). This allows us to integrate each side with respect to its variable.

Step 4: Integrate Both Sides

Integrate the left-hand side \(\int \frac{dy}{\sqrt{1-y^2}}\) which is known to give \(\sin^{-1} y + C\). Integrate the right-hand side \(\int dx = x + K\). Set the constants equal: \(C = K = c\).

Step 5: Solve for y(x)

By equating the integrated forms, we derive the equation \(\sin^{-1} y = x + c\). Solving this, we find \(y = \sin(x + c)\).

Step 6: Explore Particular Solutions

Select certain values of \(c\) to find particular solutions: - If \(c = 0\), we have \(y = \sin x\).- If \(c = \pi/2\), we get \(y = \cos x\).- Similarly, \(-y = \sin x\) implies \(y = -\sin x\), and \(-y = \cos x\) implies \(y = -\cos x\).

Step 7: Identify Singular Solutions

The equation \(y^2 + \left( \frac{dy}{dx} \right)^2 = 1\) also admits the constant solutions \(y = 1\) and \(y = -1\), since they satisfy the equation.

Key Concepts

Separable Differential Equations

A differential equation is termed separable if you can rearrange it such that all terms with one variable appear on one side of the equation, and all terms with the other variable appear on the opposite side. This makes them easier to solve with integration. In the given exercise, we start with the equation: \( \frac{dy}{dx} = \sqrt{1-y^2} \). By separating the variables, we rearrange it as \( \frac{dy}{\sqrt{1-y^2}} = dx \). This setup allows us to integrate each side separately, ultimately obtaining a general solution. Here are a few key steps to remember when working with separable differential equations:

• Rearrange terms so that one side of the equation contains only terms involving the variable \(y\) and \(dy\), and the other side contains terms involving \(x\) and \(dx\).
• Integrate both sides independently.
• Combine the constants of integration into a single constant.

Thus, separable differential equations form a foundational method for solving many problems in calculus by leveraging the process of integration.

Pythagorean Identity

The Pythagorean identity is a crucial concept when dealing with trigonometric functions. It states that for any angle \( \theta \), \( \sin^2 \theta + \cos^2 \theta = 1 \). In our problem, the equation \( y^2 + \left( \frac{dy}{dx} \right)^2 = 1 \) resembles this identity. This resemblance gives us a clue that the solutions to the differential equation might involve trigonometric functions.

• Notice how the equation mirrors the structure of \( \sin^2 \theta + \cos^2 \theta = 1 \).
• This similarity suggests that solutions involving sine and cosine functions may satisfy the initial differential equation.

Recognizing these patterns helps in identifying the potential form of the solutions. Therefore, understanding the Pythagorean identity can provide insight into solving trigonometric differential equations.

Singular Solutions

In the world of differential equations, singular solutions are those solutions which satisfy the differential equation but cannot be obtained directly from the general solution. In the given problem, besides the solutions derived using the integration of the separable equation, there are particular constant solutions: \( y = 1 \) and \( y = -1 \). These solutions do not vary with \(x\) and satisfy the equation independently:

• For \( y = 1 \), the derivative \( \frac{dy}{dx} = 0 \), thus \( 1 + 0^2 = 1 \), which satisfies the equation.
• Similarly, for \( y = -1 \), we have \( (-1)^2 + 0^2 = 1 \).

Such solutions highlight that not every solution needs to be generated from the general form derived through integration, emphasizing the uniqueness and applicability of singular solutions in differential equations.

Particular Solutions

A particular solution is a specific instance of a general solution to a differential equation, obtained by assigning specific values to the constant of integration. In the exercise provided, by choosing different values for \(c\), we can derive various particular solutions.

• If \( c = 0 \), the function \( y = \sin x \) is obtained.
• When \( c = \frac{\pi}{2} \), \( y = \cos x \) arises as a solution.
• Additionally, using \(-y\) yields \( y = -\sin x \) and \( y = -\cos x \).

By exploring the different values for \(c\), we generate different functions that all solve the differential equation given the initial condition \(c\). Particular solutions offer a precise solution tailored to specific conditions or constraints applied to the problem.