Question

Finding a Particular Solution In Exercises \(17-24,\) find the particular solution of the differential equation that satisfies the initial condition. $$ \text{Differential Equation} \quad \text{Initial Condition} $$ $$ y^{\prime}+y \sec x=\sec x \quad y(0)=4 $$

Short answer

The particular solution to the differential equation \(y'+y\sec(x)=\sec(x)\) with initial condition \(y(0)=4\) is \(y=\frac{\sec(x)+3}{\sec(x)+\tan(x)}\)

Step-by-step solution

Identify the differential equation and initial conditions

The given differential equation is \(y'+y\sec(x)=\sec(x)\) and the initial condition is \(y(0)=4\)

Rewrite the differential equation in standard form

The differential equation can be written in standard form \(y'+P(x)y=Q(x)\) as \(y' + y \sec(x)-\sec(x)=0\)

Determine the integrating factor

The integrating factor is found using the formula \(\mu(x)=e^{\int P(x) dx}\). Here, \(P(x)=\sec(x)\), thus \(\mu(x)=e^{\int \sec(x) dx} = e^{\ln|\sec(x)+\tan(x)|}= \sec(x)+\tan(x)\)

Multiply the standard-form differential equation by the integrating factor

Upon multiplying the differential equation by the integrating factor, we get \((\sec(x)+\tan(x))y'+\sec(x)(\sec(x)+\tan(x))y= \sec^2(x)+\sec(x)\tan(x)\)

Simplify and integrate both sides of the equation

The left side of the equation can be rewritten as \(( (\sec(x)+\tan(x))y)'\) and on integrating both sides with respect to \(x\) we have \(\int ((\sec(x)+\tan(x))y)' dx = \int (\sec^2(x)+\sec(x)\tan(x)) dx\), which gives \((\sec(x)+\tan(x))y = \sec(x)+ C\)

Get the general solution

Dividing through by \(\sec(x)+\tan(x)\) gives the general solution as \(y= \frac{\sec(x)+C}{\sec(x)+\tan(x)}\)

Use the initial condition to get the constant of integration and find the particular solution

Substitute \(x=0\) and \(y=4\) in the general solution expression to get that \(4= \frac{1+C}{1+0}\). Solving for \(C\) gets \(C=3\), so the particular solution is \(y=\frac{\sec(x)+3}{\sec(x)+\tan(x)}\)

Key concepts

Differential Equation

A differential equation is a mathematical equation that involves functions and their derivatives. In this exercise, we are given the differential equation: \( y' + y \sec(x) = \sec(x) \). Here, \( y' \) represents the derivative of \( y \) with respect to \( x \). Differential equations are important because they help us understand how things change. In this case, the equation is a first-order linear differential equation. The goal is to find a function \( y(x) \) that satisfies this equation. Linear differential equations can often be solved by specific methods, such as using an integrating factor, to find solutions. The exercise challenges us to use these methods to find a particular solution, given certain conditions.

Initial Condition

Initial conditions are values that specify the state of a system at a particular point. In solving differential equations, initial conditions allow us to find particular solutions that satisfy both the equation and specific conditions at a given point. For this exercise, the initial condition is \( y(0) = 4 \). This tells us the value of our function \( y \) when \( x = 0 \). With the initial conditions, we can determine the constant of integration within the context of our solution. As we solve the equation, we'll incorporate this initial condition to ensure that our solution aligns with the real-world scenario or specific case we are examining.

Integrating Factor

An integrating factor is a function used to simplify the process of solving a differential equation. The integrating factor helps transform a non-exact differential equation into an exact one, which is far easier to solve. In our exercise, we find the integrating factor using the formula \( \mu(x)=e^{\int P(x) \, dx} \). Here, \( P(x) = \sec(x) \), so the integrating factor becomes \( \mu(x) = \sec(x) + \tan(x) \). By multiplying the entire differential equation by this integrating factor, we can rearrange it into a form where the left-side resembles the derivative of a product. This is a pivotal step as it simplifies integration and aids in arriving at a general solution.

General Solution

The general solution of a differential equation represents a family of functions that include all possible solutions. For linear differential equations, the general solution includes an arbitrary constant that can take on any value. In this exercise, after applying the integrating factor, we simplify and integrate the differential equation. This process leads us to the general solution: \( y = \frac{\sec(x) + C}{\sec(x) + \tan(x)} \). The variable \( C \) is important because it accounts for all potential particular solutions of the equation. By applying initial conditions, we can determine \( C \) explicitly, thus arriving at the particular solution that satisfies both the differential equation and the initial condition: \( y = \frac{\sec(x) + 3}{\sec(x) + \tan(x)} \). This function is the specific solution that meets all given criteria in the problem.