Problem 32 - Advanced Engineering Mathematics

Question

Differentiating \(\ln \left(x^{2}+10\right)+\csc y=c\) we get \\[\begin{array}{c}\frac{2 x}{x^{2}+10}-\csc y \cot y \frac{d y}{d x}=0 \\\\\frac{2 x}{x^{2}+10}-\frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \frac{d y}{d x}=0\end{array}\\] or \\[2 x \sin ^{2} y d x-\left(x^{2}+10\right) \cos y d y=0\\] Writing the differential equation in the form \(\frac{d y}{d x}=\frac{2 x \sin ^{2} y}{\left(x^{2}+10\right) \cos y}\) we see that singular solutions occur when \(\sin ^{2} y=0,\) or \(y=k \pi,\) where \(k\) is an integer.

Step-by-Step Solution

Answer

Singular solutions occur when \( y = k\pi \) for integer \( k \).

Step 1: Differentiate the function

We are given the function \( \ln (x^2 + 10) + \csc y = c \). Differentiate both sides of the equation with respect to \( x \). The derivative of \( \ln (x^2 + 10) \) is \( \frac{2x}{x^2 + 10} \) and the derivative of \( \csc y \) with respect to \( x \) is \( -\csc y \cot y \frac{dy}{dx} \). Thus, the equation becomes \[ \frac{2x}{x^2 + 10} - \csc y \cot y \frac{dy}{dx} = 0. \]

Step 2: Substitute Trigonometric Identities

Recall the identities \( \csc y = \frac{1}{\sin y} \) and \( \cot y = \frac{\cos y}{\sin y} \). Substitute these identities into the differentiated equation: \[ \frac{2x}{x^2 + 10} - \frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \frac{dy}{dx} = 0. \]

Step 3: Rearrange the Equation

Rearrange the equation so that \( \frac{dy}{dx} \) is isolated: \[ \frac{dy}{dx} = \frac{2x \sin^2 y}{(x^2 + 10) \cos y}. \] This step involves cross-multiplying and rearranging the terms.

Step 4: Identify Singular Solutions

Singular solutions occur when the expression \( \sin^2 y \) in the numerator is equal to zero, which happens when \( \sin y = 0 \). Therefore, \( y = k\pi \), where \( k \) is an integer, represents the singular solutions of the differential equation.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of functions that are not explicitly solved for one variable in terms of another. We often encounter this scenario in equations that involve multiple variables intertwined in a single expression. Unlike explicit functions where you have something straightforward like \( y = f(x) \), implicit functions might look like \( x^2 + y^2 = 25 \). Here, y is not isolated on one side of the equation.

• Begin by differentiating each term in the implicit equation with respect to x.
• Wherever you see the variable y, treat it as a function of x (y = f(x)). So when you differentiate y, you apply the chain rule to include \( \frac{dy}{dx} \).

In the exercise provided, this was applied to the term \( \csc y \) (cosecant y). Since it was originally written with y, its differentiation with respect to x involved implicit differentiation, resulting in the term \( -\csc y \cot y \frac{dy}{dx} \). Remember, the chain rule is essential here because every differentiation involving y requires multiplying by \( \frac{dy}{dx} \). This allows us to solve for the derivative \( \frac{dy}{dx} \) when y isn't explicitly defined.

Trigonometric Identities

Trigonometric identities are fundamental in simplifying and resolving equations involving trigonometric functions. In the given solution, understanding basic trigonometric identities allowed for the simplification of the differentiated expression.Let's explore the identities used:

• \( \csc y = \frac{1}{\sin y} \): This is key for expressing cosecant in terms of sine, an easier function to differentiate and handle.
• \( \cot y = \frac{\cos y}{\sin y} \): This identity breaks down the cotangent function into sine and cosine, two basic trigonometric entities.

These identities help in rewriting complicated trigonometric expressions into simpler forms. For instance, the original expression \( \csc y \cot y \frac{dy}{dx} \) turns into \( \frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \cdot \frac{dy}{dx} \), which is simpler to manipulate and differentiate further. This transformation makes integration, differentiation, and solving equations much more manageable.

Singular Solutions

In differential equations, singular solutions are those that don’t fall within the general solution set but satisfy the differential equation itself. Singular solutions often stem from the specific conditions in the equation that set certain parts to zero, which normally wouldn’t.In the provided exercise, singular solutions were found by examining the conditions under which the equation simplifies or breaks down:

• The denominator isn't influencing the differentiation because it does not vanish to make the implicit differentiation unstable.
• The numerator’s \( \sin^2 y \) term being zero, which translates to \( \sin y = 0 \).

This represents points where the behavior of dy/dx cannot be typically described, specifically where \( y = k\pi \) and k is any integer. At these points, the usual calculation of slopes does not apply in the conventional sense, hence why they are called singular. Identifying these special solutions is crucial for fully understanding the range and behavior of the differential equation in question.