Question

In Exercises \(16-24\) find the general solution. $$ y^{\prime}+(\tan x) y=\cos x $$

Short answer

Question: Find the general solution for the first-order differential equation: \(y^{\prime}+(\tan x) y=\cos x\). Answer: The general solution for the given differential equation is: \(y(x) = (x+C)\cos x\).

Step-by-step solution

Identify the form of the given differential equation

The given differential equation has the following form: $$ y^{\prime} + p(x)y = q(x) $$ Here, \(p(x) = \tan x\) and \(q(x) = \cos x\).

Calculate the integrating factor (IF)

To calculate the integrating factor, we will use the formula: $$ IF = e^{\int p(x) dx} $$ In our case, $$ IF = e^{\int \tan x\, dx} $$ We know that \(\int \tan x\, dx = \ln |\sec x| + C\). However, we only need the continuous function \(\ln |\sec x|\). So, $$ IF = e^{\ln |\sec x|} $$ Since \(e^{\ln x} = x\), the integrating factor is: $$ IF = \sec x $$

Multiply the given equation with the integrating factor

Now, we multiply the differential equation with the integrating factor (\(\sec x\)): $$ (\sec x) (y^{\prime} + (\tan x)y) = (\sec x)(\cos x) $$ Simplifying, we get: $$ y^{\prime}\sec x + y\sec x \tan x = 1 $$

Recognize the left side as the derivative of a product

Observe that the left side of the equation in Step 3 is a derivative of the product \(y\sec x\). The derivative of \(y\sec x\) is: $$ \frac{d(y\sec x)}{dx} = y^{\prime}\sec x + y\sec x \tan x $$ So, our equation becomes: $$ \frac{d(y\sec x)}{dx} = 1 $$

Integrate both sides of the equation

Now, we integrate both sides of the equation with respect to \(x\): $$ \int \frac{d(y\sec x)}{dx}dx = \int 1 dx $$ On the left side, the integral undoes the derivative, so we get: $$ y\sec x = x + C $$

Solve for \(y\)

Finally, we solve for \(y\): $$ y = \frac{x + C}{\sec x} = (x+C)\cos x $$ Here, \(C\) is the constant of integration. So, the general solution for the given differential equation is: $$ y(x) = (x+C)\cos x $$

Key concepts

Integrating Factor Method

The integrating factor method is a powerful tool for solving first order linear differential equations. It involves finding a function, called the integrating factor (IF), which when multiplied with the original differential equation, simplifies it into an easily integrable form.

The basic formula to calculate the integrating factor is as follows:
\[\begin{equation}IF = e^{\int p(x) dx}\end{equation}\]
Where \( p(x) \) is the coefficient of \( y \) in the original differential equation of the form \( y' + p(x)y = q(x) \). The idea is that after finding the integrating factor and multiplying it across the equation, you get an expression that is the derivative of a product. This makes the equation much easier to integrate and find the general solution.

First Order Linear Differential Equation

First order linear differential equations are a class of equations that can be written in the form:
\[\begin{equation}y' + p(x)y = q(x)\end{equation}\]
Such equations consist of the first derivative of the unknown function, here denoted as \( y' \), and the function itself, \( y \), with their coefficients being functions solely of the independent variable, \( x \), here denoted as \( p(x) \) and \( q(x) \) respectively.

The solution process typically involves finding an integrating factor that transforms the original equation into a perfect derivative, which can then be integrated to find the solution. The goal of solving such equations is to find a general solution that accounts for all possible solutions, including the constant of integration, \( C \).

Integration of Trigonometric Functions

Integration of trigonometric functions is a frequent requirement when solving differential equations. It involves finding the antiderivatives of trigonometric expressions.

For example, in our exercise, we encounter the integration of \( \tan x \), which results in \( \ln |\sec x| + C \). However, when using this result as part of the integrating factor in differential equations, the constant \( C \) is unnecessary because we are looking for a function that will simplify the equation.

It's important to note that integrating trigonometric functions often utilizes known integral identities, substitutions, or transformations to simplify the process. Familiarity with these techniques is essential for efficiently solving integrals involving trigonometric functions.