Problem 14 - Thomas Calculus

Question

Solve the differential equations. \(\tan \theta \frac{d r}{d \theta}+r=\sin ^{2} \theta, \quad 0<\theta<\pi / 2\)

Step-by-Step Solution

Answer

The general solution is \( r(\theta) = \frac{1}{2} \sin^2 \theta + C \sin \theta \).

Step 1: Separate Variables

The differential equation is given as \( \tan \theta \frac{d r}{d \theta}+r=\sin ^{2} \theta \). We can start by writing \( \tan \theta \) as \( \frac{\sin \theta}{\cos \theta} \), to get \( \frac{\sin \theta}{\cos \theta} \frac{d r}{d \theta} + r = \sin^2 \theta \). This makes it easier to manipulate the equation for solving.

Step 2: Rewrite the Equation

Rearrange the equation to isolate the derivative term: \( \frac{\sin \theta}{\cos \theta} \frac{d r}{d \theta} = \sin^2 \theta - r \). Multiply through by \( \cos \theta \) to get: \( \sin \theta \frac{d r}{d \theta} = \sin^2 \theta \cos \theta - r \cos \theta \).

Step 3: Simplify and Rearrange

Factor out \( \sin \theta \) on the left: \( \frac{d r}{d \theta} = \frac{\sin^2 \theta \cos \theta}{\sin \theta} - \frac{r \cos \theta}{\sin \theta} \). This becomes \( \frac{d r}{d \theta} = \sin \theta \cos \theta - r \cot \theta \).

Step 4: Introduce Substitution

Use substitution to simplify further, let \( r = u \sin \theta \). Then, take the derivative: \( \frac{d r}{d \theta} = u \cos \theta + \frac{du}{d\theta} \sin \theta \). Substitute back: \( u \cos \theta + \frac{du}{d\theta} \sin \theta = \sin \theta \cos \theta - u \cos \theta \).

Step 5: Cancel Out Terms

After substitution, cancel like terms: \( \frac{du}{d\theta} \sin \theta = \sin \theta \cos \theta - u \cos \theta \), leading to \( \frac{du}{d\theta} = \cos \theta - u \cot \theta \). Here, \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).

Step 6: Solve for \( u \) using Separable Variables

Rewrite the equation as \( \frac{du}{d\theta} + u \cot \theta = \cos \theta \). Move \( u \cot \theta \) to the right side: \( \frac{du}{d\theta} = \cos \theta - u \cot \theta \). Use the integrating factor method: Integrating factor = \( e^{\int \cot \theta \, d\theta} = \sin \theta \).

Step 7: Integrate Both Sides

Multiply throughout by the integrating factor: \( \sin \theta \frac{du}{d\theta} + u \cos \theta = \sin \theta \cos \theta \). Integrate both sides: \( \frac{d}{d\theta}(u \sin \theta) = \sin \theta \cos \theta \). Upon integration, \( u \sin \theta = \frac{1}{2} \sin^2 \theta + C \).

Step 8: Substitute Back for \( r \)

Recall the substitution \( r = u \sin \theta \). Substitute back to get \( r = \frac{1}{2} \sin^2 \theta + C \sin \theta \).

Step 9: General Solution

The general solution is \( r(\theta) = \frac{1}{2} \sin^2 \theta + C \sin \theta \), for some constant \( C \).

Key Concepts

Separation of Variables

Separation of variables is a powerful technique for solving differential equations. This method is particularly useful when you can rewrite the equation such that each variable and its derivative are on separate sides of the equation.
To begin using separation of variables, identify the parts of the equation that can be separated cleanly. For instance, if you have an equation like \( \frac{dy}{dx} = g(x)h(y) \), you would rearrange it to \( \frac{1}{h(y)} dy = g(x) dx \).
This method involves the following steps:

• Rewriting the equation such that all instances of one variable and its derivative are on one side, and all instances of the other variable are on the other side.
• Integrating both sides with respect to their relevant variables.
• Using the initial conditions or additional information if provided, to solve for the integration constant.

Remember that the goal is to end up with a function or expression where each term only involves one variable, making the integration process straightforward.

Integrating Factor Method

The integrating factor method is essential when dealing with first-order linear differential equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). The key to this method is finding a function, known as the integrating factor, which simplifies the integration process.
First, you calculate the integrating factor using the formula \( e^{\int P(x) \, dx} \). This function transforms the equation into an exact differential that can be integrated directly.
Steps for this process include:

• Identify and write down the equation in the standard form.
• Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply every term in the differential equation by \( \mu(x) \) to make the left-hand side an exact derivative \( \frac{d}{dx}[\mu(x)y] = \mu(x)Q(x) \).
• Integrate both sides with respect to \( x \).
• Solve for \( y \) and apply initial conditions if provided.

This method effectively reduces the problem to a simple integration task, which can then be solved to find the particular solution of the differential equation.

Trigonometric Substitution

Trigonometric substitution is a technique often used to simplify integrals, but it can also help simplify differential equations by exploiting trigonometric identities. This substitution uses relationships among trigonometric functions to convert complex expressions into more manageable forms.
For instance, common substitutions include using \( \sin \theta, \cos \theta, \) or \( \tan \theta \) in place of certain variable expressions. These substitutions often involve Pythagorean identities, like \( \sin^2 \theta + \cos^2 \theta = 1 \), which can help reduce complex terms.
When applying trigonometric substitution, follow these steps:

• Identify expressions in the equation that can be replaced with trigonometric identities, such as recognizing \( \sqrt{a^2 - x^2} \) as \( a \sin \theta \).
• Rewrite the entire equation in terms of the trigonometric functions you've chosen.
• Simplify using known trigonometric identities.
• Continue solving the equation as usual, often leading to simpler integrations.

Trigonometric substitutions are particularly useful for integrals or differential equations involving square roots and can greatly simplify the algebraic manipulation required to reach a solution.