Problem 9 - Calculus: Early Transcendentals

Question

Find the general solution of the equation. $$y^{\prime} \cos t+y \sin t=1,-\pi / 2

Step-by-Step Solution

Answer

The general solution is \( y(t) = \ln | \sec t + \tan t | + C \cos t \).

Step 1: Identify the Type of Differential Equation

The given equation is \( y' \cos t + y \sin t = 1 \). This is a first-order linear differential equation. It can be written in the standard form \( y' + P(t) y = Q(t) \) where \( P(t) = \tan(t) \) and \( Q(t) = \sec(t) \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(t) \) is given by \( \mu(t) = e^{\int P(t) \, dt} \). Substitute \( P(t) = \tan(t) \), so \( \mu(t) = e^{\int \tan(t) \, dt} = e^{\ln|\sec t|} = |\sec t| \). Since \( -\pi/2 < t < \pi/2 \), \( \sec t \) is always positive, thus \( \mu(t) = \sec t \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \( \sec t \):\[ \sec t \cdot (y' \cos t + y \sin t) = \sec t \cdot 1 \]This simplifies to \( y' + y \tan t = \sec t \).

Step 4: Simplify the Left-Hand Side to Exact Derivative

The left-hand side of the equation \( y' + y \tan t \) can be expressed as the derivative of \( y \sec t \) because:\[ \frac{d}{dt}(y \sec t) = y' \sec t + y (\sec t \tan t) = y' \cos t + y \sin t \].

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(y \sec t) \, dt = \int \sec t \, dt \]. This leads to:\( y \sec t = \ln | \sec t + \tan t | + C \), where \( C \) is the arbitrary constant.

Step 6: Solve for y

To find \( y(t) \), solve:\[ y \sec t = \ln | \sec t + \tan t | + C \]Thus, \( y(t) = \sec t \cdot (\ln | \sec t + \tan t | + C) \).However, for simplicity, multiply both sides by \( \cos t \) to get:\( y(t) = \ln | \sec t + \tan t | + C \cos(t) \).

Key Concepts

Integrating Factor

In solving first-order linear differential equations, the integrating factor is a handy tool. It simplifies the process of finding solutions by turning the equation into something that can be directly integrated. Imagine it as a magical function that, when multiplied across the equation, aligns the terms just right!

Here's how it works: For an equation in the form \[ y' + P(t) y = Q(t), \]we derive the integrating factor \( \mu(t) \) using the formula:\[\mu(t) = e^{\int P(t) \, dt}\]The integrating factor essentially "uncovers" the underlying structure of the equation.

• In this case, our \(P(t)\) is \(\tan(t)\).
• Thus, after integrating \(\tan(t)\), we arrive at the integrating factor \(\mu(t) = \sec(t)\).

This \(\mu(t)\) helps you reorganize the equation into a form that could be expressed as the derivative of a product, making the integration super straightforward.

General Solution

The concept of a general solution is akin to finding a "universal key" for solving differential equations. Once you arrive at this solution, it includes arbitrary constants, allowing it to fit a wide array of specific cases or initial conditions.
When dealing with the given differential equation, we've already employed the integrating factor. By multiplying the whole equation by \(\sec(t)\), it was reshaped into a derivative form:\[\frac{d}{dt}(y \sec t) = \sec t\]Upon integrating both sides, we arrive at:\[y \sec t = \ln|\sec t + \tan t| + C\]where \(C\) represents the arbitrary constant.

To express \(y\) explicitly, multiply through by \(\cos t\) to isolate \(y(t)\). Hence, the general solution takes the form:\[y(t) = \ln|\sec t + \tan t| + C \cos(t)\] This expression covers all possible solutions to the equation over the interval \(-\pi/2 < t < \pi/2\), illustrating how a solution adjusts based on different initial conditions.

Trigonometric Functions

Trigonometric functions are fundamental in this exercise, as they serve as both coefficients and solutions in differential equations. They describe periodic phenomena, such as cycles or waves, which is critical when examining models of natural phenomena.
In this particular exercise, recognizing trigonometric identities and their properties was crucial:

• We used \(\cos(t)\) and \(\tan(t)\) as parts of our differential equation.
• \(\sec(t)\) made a significant appearance as both the inverse of \(\cos(t)\) and our integrating factor
  throughout the solution.

Trigonometric identities also provide insights that simplify integration and differentiation processes. Equipped with knowledge of sine, cosine, tangent, and secant functions, we are better prepared to manipulate and solve equations that are woven into the fabric of periodic and oscillatory functions.