Problem 25 - Advanced Engineering Mathematics

Question

Let \(u=x+y\) so that \(d u / d x=1+d y / d x .\) Then \(\frac{d u}{d x}-1=\tan ^{2} u\) or \(\cos ^{2} u d u=d x .\) Thus \(\frac{1}{2} u+\frac{1}{4} \sin 2 u=x+c\) or \(2 u+\sin 2 u=4 x+c_{1},\) and \(2(x+y)+\sin 2(x+y)=4 x+c_{1}\) or \(2 y+\sin 2(x+y)=2 x+c_{1}\).

Step-by-Step Solution

Answer

The solution expresses as \( 2(x+y) + \sin 2(x+y) = 4x + c_1 \) or \( 2y + \sin 2(x+y) = 2x + c_1 \).

Step 1: Identify Key Derivative and Substitution

Start with the given substitution, let \( u = x + y \). Given that \( \frac{du}{dx} = 1 + \frac{dy}{dx} \), this implies that the change in \( u \) with respect to \( x \) is dependent on the derivative of \( y \) with respect to \( x \).

Step 2: Rearrange the Derivative Equation

From \( \frac{du}{dx} = 1 + \frac{dy}{dx} \), rearrange to find an expression for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{du}{dx} - 1 \).

Step 3: Recognize Relationship with Trig Function

Given \( \frac{du}{dx} - 1 = \tan^2 u \), understand that this equation equates the derivative of \( u \) adjusted by subtracting 1 to the square of the tangent of \( u \).

Step 4: Integrate Using Trigonometric Identity

Integrate the equation \( \cos^2 u \, du = dx \), knowing that \( 1 - \tan^2 u = \cos^2 u \) from the trigonometric identity. This gives a relationship between \( u \) and \( x \).

Step 5: Solve the Integrated Equation

By integrating \( \cos^2 u \, du = dx \), we obtain \( \frac{1}{2} u + \frac{1}{4} \sin 2u = x + c \). This is from integrating \( \cos^2 u \), which results in this specific antiderivative form involving \( u \).

Step 6: Simplify to Achieve Final Expression

By multiplying through by 4, we achieve \( 2u + \sin 2u = 4x + c_1 \). This simplifies the integration result into a more compact form using \( c_1 \) as a constant of integration.

Step 7: Substitute Back to Original Variables

Substitute back \( u = x + y \) to form \( 2(x+y) + \sin 2(x+y) = 4x + c_1 \), which represents the relationship in terms of the original variables \( x \) and \( y \).

Step 8: Rearrange for Final Equation in Terms of y

Rearrange the equation to isolate \( y \): \( 2y + \sin 2(x+y) = 2x + c_1 \). This is the final form expressed in terms of the original variables.

Key Concepts

Trigonometric Identities

Trigonometric identities, such as those involving sine, cosine, and tangent, play a crucial role in solving differential equations involving trigonometric functions. In the exercise provided, knowing that

• \( 1 - \tan^2 u = \cos^2 u \)

helps us decipher how changes in the function relate to one another. This identity is derived from the well-known Pythagorean identity \( \sin^2 u + \cos^2 u = 1 \), reorganizing it to express tangent and cosine in terms of each other.
Recognizing such identities simplifies integration and differentiation involving trigonometric functions. Use these identities to help transform and integrate expressions by substituting known identities or by rearranging expressions to match known forms.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse differentiation. They can be tough to find without practice, especially when trigonometric functions are involved. In the exercise, we encounter an antiderivative of the form:

• \( \frac{1}{2} u + \frac{1}{4} \sin 2u = x + c \)

This result comes from the integration of \(\cos^2 u\). Knowing how to handle trigonometric integrations and their antiderivatives is vital. Always remember

• \( \int \sin u \, du = -\cos u + C \)
• \( \int \cos u \, du = \sin u + C \)

where C is the constant of integration. In practical terms, finding the antiderivative allows you to backtrack from an observed rate of change back to the original quantity.

Substitution Method

The substitution method is a powerful technique for simplifying complex integrals by introducing a new variable. This exercise demonstrates substitution by setting

• \( u = x + y \)

This simplifies the relationship into one variable. Here, the derivative \( \frac{du}{dx} = 1 + \frac{dy}{dx} \) shows how changes in \( u \) relate to changes in \( x \) and \( y \). Instead of handling a complicated differential directly, use the substitution to work within simpler terms.
When performing substitution, ensure to substitute both the variable and the differential part of the integral. This yields simpler equations, making them easier to solve or integrate.

Integration Techniques

Various integration techniques aid in solving integrals involving complex or non-standard forms. In this exercise, the integral of \( \cos^2 u \) is tackled. Understanding integration formulas and when to apply them simplifies calculations. Common techniques include:

• Partial fraction decomposition
• Substitution
• Integration by parts

For trigonometric functions:

• Recognize identities and transform complex forms into integrable ones

Identify patterns or relationships, often using trigonometric identities, to transform an expression into an easier form. Breaking down a complex function into elementary parts using substitution or applying standard integration rules is critical.
When integrating something like \( \cos^2 u \), use the identity \( \cos^2 u = \frac{1}{2}(1 + \cos(2u)) \) to simplify the integration process. This approach highlights the need for a broad familiarity with trigonometric identities and the ability to recognize which integration technique to apply.