Question

Solve the initial-value problems. Solve each of the following equations of the form (2.41): (a) \(\cos y \frac{d y}{d x}+\frac{1}{x} \sin y=1\). (b) \((y+1) \frac{d y}{d x}+x\left(y^{2}+2 y\right)=x\).

Short answer

\(y(x) = e^{-\int \frac{1}{x} dy} \left( \int \frac{e^{\int \frac{1}{x} dy}}{\cos{y}} dx + C\right)\)

Step-by-step solution

Analyze the differential equation (a)

In equation (a), the given differential equation is \(\cos y \frac{d y}{d x}+\frac{1}{x} \sin y=1.\) We notice this is a first-order linear homogeneous differential equation in terms of x and y. Following the steps below to find the general solution.

Identify the integrating factor of the differential equation (a)

The differential equation can be expressed in the linear form: \(\frac{dy}{dx} + \frac{\sin{y}}{x \cos{y}} = \frac{1}{\cos{y}}\) Comparing with the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where, for this case \(P(x) =\frac{\sin{y}}{x \cos{y}} \) and \(Q(x)=\frac{1}{\cos{y}}\). Our first step is to find the integrating factor, which is given by \[IF(x) = e^{\int P(x) dx}.\] Since P(x) in our case involves y as well, we should work with respect to y: \(P(y)=\frac{1}{x}\), so the integrating factor IF(y) can be written as: IF(y) = \(e^{\int P(y) dy} = e^{\int \frac{1}{x} dy}.\)

Multiply both sides of the differential equation (a) by integrating factor

Now we multiply both sides of our linear equation by the integrating factor \(e^{\int \frac{1}{x} dy}\) to obtain a new differential equation: \(\frac{d}{dx}\left(e^{\int \frac{1}{x} dy} y\right) = \frac{e^{\int \frac{1}{x} dy}}{\cos{y}}\)

Integrate RHS and find the general solution of the differential equation (a)

Now, we integrate both sides with respect to x, \(\int \frac{d}{dx}\left(e^{\int \frac{1}{x} dy} y\right) dx = \int \frac{e^{\int \frac{1}{x} dy}}{\cos{y}} dx\) On integrating, we get \(e^{\int \frac{1}{x} dy} y = \int \frac{e^{\int \frac{1}{x} dy}}{\cos{y}} dx + C\) Now we find the general solution for y,\[y(x) = e^{-\int \frac{1}{x} dy} \left( \int \frac{e^{\int \frac{1}{x} dy}}{\cos{y}} dx + C\right)\]

Key concepts

First-Order Differential Equations

Understanding first-order differential equations is essential to grasp more complex mathematical concepts. They are a type of differential equation that involves only the first derivative of the unknown function and the function itself. The general form is written as \( \frac{dy}{dx} = f(x, y) \) where\( x \) and \( y \) are variables, and \( f(x, y) \) is a given function.

These equations are the stepping stones into the world of calculus and appear in various applications such as physics, engineering, and economics. The key to solving these equations often involves separating variables, using an integrating factor, or special methods if the differential equation fits a particular type. In the given exercise, the differential equation \(\cos y \frac{d y}{d x}+\frac{1}{x} \sin y=1\) is a first-order equation that is solved using an integrating factor.

Integrating Factor

The integrating factor is a powerful tool for solving linear first-order differential equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\). It is essentially a function, usually denoted as \( IF(x) \) or \( IF(y) \), that when multiplied to both sides of the differential equation, transforms it into an easily integrable form.

Typically, an integrating factor is expressed as \(\text{IF}(x) = e^{\int P(x) dx}\), and its main role is to allow the left side of the differential equation to be written as the derivative of a product.

How to Use Integrating Factor

• Identify the function \( P(x) \) from the differential equation.
• Calculate the integrating factor \( e^{\int P(x) dx} \).
• Multiply every term in the differential equation by the integrating factor.
• Upon multiplication, the left side should represent the derivative of a product, which can then be integrated to find a solution.

In the provided step-by-step solution, the correct integrating factor is used to successfully transform the equation into a version that is solvable through integration.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a specific type of differential equation that, when set to zero, does not have any terms that are independent of the unknown function and its derivatives. In simpler terms, every term in such an equation involves either the function itself or one of its derivatives. The general form of a linear homogeneous first-order differential equation is \(\frac{dy}{dx} + P(x)y = 0\).

The term 'homogeneous' refers to the absence of a non-zero function \( Q(x) \) on the right-hand side of the equation. Such equations are considered simpler to solve than their non-homogeneous counterparts because they can often be transformed into a separable equation by finding an appropriate substitution.

Characteristics and Solutions

• Solutions to homogeneous equations exhibit properties like superposition, meaning the sum of two solutions is also a solution.
• The general solution often involves the exponential function resulting from integrating the function \( P(x) \).

In the exercise given, the equation initially appears non-homogeneous but is manipulated into a form that allows the use of an integrating factor, and hence, can effectively be solved as a linear equation.