Problem 29

Question

Let $u=x+y$ so that $d u / d x=1+d y / d x .$ Then $\frac{d u}{d x}-1=\cos u$ and $\frac{1}{1+\cos u} d u=d x .$ Now $$\frac{1}{1+\cos u}=\frac{1-\cos u}{1-\cos ^{2} u}=\frac{1-\cos u}{\sin ^{2} u}=\csc ^{2} u-\csc u \cot u$$ so we have $\int\left(\csc ^{2} u-\csc u \cot u\right) d u=\int d x$ and $-\cot u+\csc u=x+c .$ Thus $-\cot (x+y)+\csc (x+y)=x+c$ Setting $x=0$ and $y=\pi / 4$ we obtain $c=\sqrt{2}-1 .$ The solution is $$\csc (x+y)-\cot (x+y)=x+\sqrt{2}-1.$$

Step-by-Step Solution

Verified

Answer

The solution is $ \csc(x+y) - \cot(x+y) = x + \sqrt{2} - 1 $.

1Step 1: Differentiate u with Respect to x

Given that $ u = x + y $, differentiate both sides with respect to $ x $. This gives $ \frac{du}{dx} = 1 + \frac{dy}{dx} $.

2Step 2: Formulate the Differential Equation

Rearrange the given differential equation $ \frac{du}{dx} - 1 = \cos u $ into a more manageable form. From Step 1, substitute $ \frac{du}{dx} = 1 + \frac{dy}{dx} $, which implies $ \frac{du}{dx} = \cos u + 1 $.

3Step 3: Integrate Both Sides

Integrate both sides of the equation: $ \int \frac{1}{1 + \cos u} du = \int dx $. To solve this integral, use trigonometric identities to rewrite it.

4Step 4: Use Trigonometric Identities

Express $ \frac{1}{1 + \cos u} $ as $ \frac{1 - \cos u}{\sin^2 u} = \csc^2 u - \csc u \cot u $. This simplifies the integration process.

5Step 5: Perform the Integrals

Integrate: $ \int ( \csc^2 u - \csc u \cot u) du $ results in $ -\cot u + \csc u = x + C $.

6Step 6: Solve for Constant C

Substitute $ x = 0 $ and $ y = \frac{\pi}{4} $ to determine $ C $. Solving $ -\cot(\frac{\pi}{4}) + \csc(\frac{\pi}{4}) = \sqrt{2} - 1 $, we find $ C = \sqrt{2} - 1 $.

7Step 7: State the General Solution

The complete solution to the differential equation is $ \csc(x+y) - \cot(x+y) = x + \sqrt{2} - 1 $. This represents the relationship between the variables based on the manipulated and integrated form of the original equations.

Key Concepts

Integration TechniquesTrigonometric IdentitiesInitial Conditions

Integration Techniques

Integration is a fundamental technique in solving differential equations. It helps us find the function that corresponds to the rate of change depicted by the differential equation. In this exercise, we are tasked with integrating both sides of the equation:

Before integrating, the given equation is manipulated into the form: $ \int \frac{1}{1 + \cos u} \, du = \int dx $.
The integral $ \int dx $ is straightforward and results in $ x + C $, where $ C $ is the constant of integration.

Integrating $ \frac{1}{1+\cos u} $ is not as direct and requires the application of trigonometric identities to simplify it into a form that is easier to integrate.
Once it is simplified to $ \int ( \csc^2 u - \csc u \cot u) du $, the integration becomes manageable. We find that:

$ \int \csc^2 u \, du = -\cot u + C $ (since the derivative of $ \cot u $ is $ -\csc^2 u $)
$ \int \csc u \cot u \, du = -\csc u + C $ (since the derivative of $ \csc u $ is $ -\csc u \cot u $)

The combined integration results from these parts give us $ -\cot u + \csc u = x + C $, forming the heart of our solution.

Trigonometric Identities

Trigonometric identities play an essential role in manipulating and simplifying functions involved in differential equations. These identities allow us to express trigonometric ratios in alternate forms, which can often simplify integration. In the exercise:

The term $ \frac{1}{1+\cos u} $ is daunting at first, but using the identity for the difference of squares $ 1 - \cos^2 u = \sin^2 u $, it's rewritten as $ \frac{1-\cos u}{\sin^2 u} $.
Recognizing that $ \csc u = \frac{1}{\sin u} $ and $ \cot u = \frac{\cos u}{\sin u} $, the expression $ \frac{1-\cos u}{\sin^2 u} $ transforms into $ \csc^2 u - \csc u \cot u $.

By utilizing these identities, the integration becomes much more feasible because it aligns with known derivative formulas of trigonometric functions. Understanding and applying these identities is crucial for efficiently handling and solving such integrals in calculus.

Initial Conditions

Initial conditions are necessary to find particular solutions to differential equations, especially when an arbitrary constant $ C $ appears from integration. They serve as specific values that let us solve for this constant, which tailors the general solution to satisfy these specific criteria.

In this problem, the initial conditions are given as $ x = 0 $ and $ y = \frac{\pi}{4} $. These are substituted back into the general solution $ -\cot(x + y) + \csc(x + y) = x + C $ to solve for $ C $.
Substituting these values results in the equation $ -\cot(\frac{\pi}{4}) + \csc(\frac{\pi}{4}) = \sqrt{2} - 1 $, revealing $ C $ as $ \sqrt{2} - 1 $.

So why does this matter? Well, the constant $ C $ adjusts the general solution to a specific context, making it applicable to the scenario described by the initial conditions. This process ensures that our solution is not only theoretically correct but also practically useful in modeling real-world phenomena.

Problem 29

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