Problem 17 - Calculus: Early Transcendentals

Question

Find the solution of the differential equation that satisfies the given initial condition. \( y' \tan x = a + y, y(\pi/3) = a, 0 < x < \pi/2 \)

Step-by-Step Solution

Answer

The solution is \( y(x) = a(4\cos x - 1) \).

Step 1: Rearrange the Equation

Start with the given differential equation: \( y' \tan x = a + y \). To make it separable, rearrange as follows: \( y' = \frac{a + y}{\tan x} \).

Step 2: Substitute to Make the Equation Separable

Substitute \( u = y + a \). Then, \( y = u - a \) and the differential \( dy = du \). The equation becomes \( \frac{du}{\tan x} = u \frac{1}{\tan x} \).

Step 3: Separate Variables

Separate variables to isolate each: \( \frac{1}{u} du = \tan x \, dx \). Integrate both sides: \( \int \frac{1}{u} \, du = \int \tan x \, dx \).

Step 4: Integrate Both Sides

The left side integrates to \( \ln |u| + C_1 \), while the right side is \( -\ln |\cos x| + C_2 \). Thus, \( \ln |u| = -\ln |\cos x| + C_2 \).

Step 5: Solve for \( u \)

Exponentiate both sides to solve for \( u \): \( |u| = e^{C_2} \cdot |\cos x| \). We can remove the absolute values because \( u = y + a \) is positive given the initial conditions and range, thus \( u = C \cdot \cos x \) for some constant \( C \).

Step 6: Apply Initial Condition

Use the initial condition \( y(\pi/3) = a \) to find \( C \). At \( x = \pi/3 \), \( u = a + a = 2a \) and \( \cos(\pi/3) = \frac{1}{2} \), so \( 2a = C \cdot \frac{1}{2} \). Solving gives \( C = 4a \).

Step 7: Substitute Back to Find \( y \)

Substitute back to find \( y \): \( y = u - a = 4a \cos x - a \). This simplifies to \( y = 4a \cos x - a \).

Step 8: Write the General Solution

Finally, the solution of the differential equation given the initial condition is \( y(x) = a(4\cos x - 1) \). This satisfies both the differential equation and the initial condition.

Key Concepts

Separable Differential Equations

Separable differential equations are a special type of differential equation. They are called "separable" because we can rearrange the variables so that each side of the equation only contains one type of variable. This makes them easier to solve. In our exercise, we started with the differential equation \( y' \tan x = a + y \). The first step was to rearrange it as \( y' = \frac{a + y}{\tan x} \).
This rearrangement allowed us to separate the variables. By substituting \( u = y + a \), we transformed the equation into \( \frac{du}{u} = \tan x \, dx \). Now, each side has one variable, making it ready for integration. This separation is crucial in simplifying the solving process of these types of equations.

Initial Conditions

Initial conditions in differential equations specify an exact solution. They are values given for the function at a certain point, allowing us to determine the constants of integration that appear after solving the differential equation. In the exercise, the initial condition given is \( y(\pi/3) = a \).
This tells us the specific value of the function \( y \) when \( x = \pi/3 \). It plays a key role in finding the specific constant \( C \).
By applying the initial condition in step 6 of the solution, we determined that \( C = 4a \), a crucial part of obtaining the final solution. Initial conditions help zero in on one specific solution among the family of possible solutions.

Integration Techniques

Integration is a fundamental technique for solving differential equations, especially separable types. Once we separated the variables, we needed to integrate both sides of the equation. The left side of the equation \( \int \frac{1}{u} \, du \) is a standard integral that results in \( \ln |u| \).
The right side, \( \int \tan x \, dx \), required recognizing it as \( \int \frac{\sin x}{\cos x} \, dx \). One way to solve this is by substitution, letting \( v = \cos x \), which simplifies the integration.
These techniques showcase how integration transforms the separated differential equation into solvable expressions, leading us closer to the solution that satisfies the given initial conditions.

Trigonometric Identities

Trigonometric identities simplify and solve parts of equations involving trigonometric functions. In this exercise, knowing the identity \( \int \tan x \, dx = -\ln |\cos x| \) was essential in integrating the right side of our separated differential equation.
Moreover, using \( \cos(\pi/3) = \frac{1}{2} \) allowed us to apply the initial condition effectively. These identities take complex trigonometric expressions and simplify them, making integration possible.
Trigonometric identities are like shortcuts in calculus. They help simplify parts of equations, making it easier to work through problems involving trigonometric functions and their derivatives.