Question

Solving a First-Order Linear Differential Equation In Exercises \(5-14,\) solve the first-order linear differential equation. $$ (y+1) \cos x d x-d y=0 $$

Short answer

The general solution to the differential equation \((y+1) \cos x d x-d y=0\) would be determined by integrating the equation \(1\cdot dy = -\cos(x)(y + 1) dx \).

Step-by-step solution

Write the Differential Equation in Standard Form

First, rewrite the given differential equation in the form of \(dy/dx + P(x)y = Q(x)\). This can be done via rearranging and a simple division operation: \(dy/dx = - \frac{(y+1)\cos(x)}{1}\)

Identify P(x) and Q(x)

P(x) is the function multiplying y in our differential equation, and Q(x) is the function on the right side of the equation. Thus we have the following: \(P(x) = 0\) and \(Q(x) = -\cos(x)(y + 1)\)

Find the Integrating Factor

The integrating factor for the differential equation is given by \(e^{∫P(x)dx}\), Since \(P(x) = 0\), the integrating factor is 1.

Multiply the entire equation by the integrating factor

Multiplying the entire equation by the integrating factor, we get \[1 \cdot \frac{dy}{dx} = -1 \cdot \cos(x)(y + 1) \]

Solve the Linear Differential Equation

The right-hand side is now an exact derivative. We can solve the resulting differential equation through integration. So the solution y(x) can be found by integrating the equation \(1\cdot dy = -\cos(x)(y + 1) dx \)

Key concepts

First-Order Differential Equation

A first-order differential equation is a relationship involving a function and its first derivative. It is called 'first-order' because it involves only the first derivative of the function, not any higher derivatives. This makes it simpler than higher-order differential equations.
An example of such an equation is \[ \frac{dy}{dx} + P(x)y = Q(x) \]where \( y \) is the function we want to find, and \( \frac{dy}{dx} \) is its first derivative.
These equations frequently appear in real-world problems involving growth and decay processes, such as population dynamics, or in electrical circuits with resistors and capacitors.

Integrating Factor

When dealing with a first-order linear differential equation, one of the efficient methods for solving it is using an integrating factor. An integrating factor is a function that, when multiplied through the differential equation, helps transform it into a form that is easier to solve.
The integrating factor is typically denoted as \( \mu(x) \) and is calculated using the formula:\[ \mu(x) = e^{\int P(x)dx} \]
This factor simplifies the solution process by transforming the left-hand side of the equation into the derivative of a product, making it more straightforward to integrate.

• For example, if \( P(x) = 0 \), then \( \mu(x) = e^{\int 0 dx} = 1 \). This means the equation becomes easier, as the multiplying factor does not change the equation.

Standard Form of Differential Equation

To solve a first-order linear differential equation, it must first be written in its standard form. This typically looks like:\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Getting to this form often involves rearranging terms, and it allows us to easily identify the functions \( P(x) \) and \( Q(x) \), which are crucial for finding the integrating factor.
Consider the example:\[ (y+1) \cos x dx - dy = 0 \]By rearranging, it becomes:\[ \frac{dy}{dx} = - (y+1)\cos(x) \]From here, we can identify that:

• \( P(x) = 0 \)
• \( Q(x) = -\cos(x)(y + 1) \)

This form is critical as it sets up the next steps, particularly involving the integrating factor.

Solving Differential Equations

Once we have the differential equation in standard form and the integrating factor identified, the next step is solving the equation. This involves integrating both sides of the equation to find the function \( y(x) \).
In the example given:\[ 1 \cdot \frac{dy}{dx} = -1 \cdot \cos(x)(y + 1) \]
Here, the equation is ready to be integrated directly as the right-hand side is an exact derivative. Solving it involves integrating both sides:

• Perform the integration on both sides to find the solution \( y(x) \).

This method lets us find the general solution to the differential equation, giving us insight into how the function \( y \) behaves over the domain of \( x \). This technique is powerful for tackling a wide range of real-world mathematical problems.