Question

Prove that if the family of integral curves of the differential equation $$\frac{d y}{d x}+p(x) y=q(x), \quad p(x) \cdot q(x) \neq 0$$ is cut by the line \(x=k,\) the tangents at the points of intersection are concurrent.

Short answer

The tangents at the points of intersection of the family of integral curves of the given differential equation \(\frac{dy}{dx} + p(x)y = q(x)\) and the line \(x=k\) are concurrent.

Step-by-step solution

Define the Differential Equation

The given differential equation is a first-order linear differential equation of the form \(\frac{dy}{dx} + p(x)y = q(x)\) where \(p(x)\) and \(q(x)\) are not equal to 0.

General Solution of the Differential Equation

This is a linear differential equation, hence the solution can be given by the method of integrating factor. The general solution is given by \(y(x) = \frac{1}{u(x)}\int u(x)q(x)dx\) where \(u(x) = \exp(\int p(x)dx)\) is the integrating factor.

Find the Derivative of the General Solution

To determine the tangents at the points of intersections, we need to find the derivative of the general solution i.e \(\frac{dy}{dx}\). Apply the product rule on the general solution to find the derivative.

Verify the Concurrence of the Tangents

Verify that the tangents are concurrent by showing that their point of intersection does not depend on the particular solution chosen. The tangents are concurrent if they intersect at the same point for all integral curves.

Key concepts

Understanding First-Order Linear Differential Equations

First-order linear differential equations are fundamental in the study of differential equations. These equations have the general form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) only.
Ease of understanding comes first from recognizing that 'linear' suggests that the variable \( y \) and its derivative \( \frac{dy}{dx} \) appear to the first power and are not multiplied together. Solving such equations often involves finding the value of \( y \) as a function of \( x \), given the functions \( P(x) \) and \( Q(x) \).
These equations frequently model real-world phenomena such as growth processes, cooling systems, and electrical circuits, emphasizing the practicality of understanding how to approach them.

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear differential equations. It starts by multiplying the entire differential equation by an integrating factor, a specially chosen function \( u(x) \), designed to make the left-hand side of the differential equation a product rule expression.
The integrating factor is generally given by the exponential function of the integral of \( P(x) \): \( u(x) = e^{\int P(x)dx} \).
Once the equation is multiplied by the integrating factor, it can be written as \( \frac{d}{dx}[u(x)y] = u(x)Q(x) \).
This allows us to integrate both sides with respect to \( x \) and find the general solution for \( y \). The integration not only simplifies the equation but with the right-hand side now easier to integrate, it paves the way to finding the particular solution.

General Solution of Differential Equations

The general solution of a differential equation represents the comprehensive set of possible solutions, accounting for all arbitrary constants. For the first-order linear equation, after applying the integrating factor, the general solution is \( y(x) = \frac{1}{u(x)} \int u(x)Q(x)dx \) plus an arbitrary constant \( C \).
This general solution encompasses all integral curves that graphically represent the function \( y(x) \). In particular, the general solution to our exercise includes the different curve positions for each value of the constant \( C \). Understanding the general solution's formation echoes how a family of solutions can derive from a single equation, embodying the essence of calculus' interpretative power in graphical analysis.

Applying the Product Rule in Calculus

The product rule in calculus is a procedure used to differentiate functions that are products of two more basic functions. The rule states \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \), where \( u \) and \( v \) are functions of \( x \).
This rule is essential when finding the derivative of the general solution in our exercise. The derivative \( \frac{dy}{dx} \) of the general solution involves applying the product rule to two functions: \( \frac{1}{u(x)} \) and the integral \( \int u(x)Q(x)dx \).
Understanding and applying the product rule is crucial because it allows for the derivative of the general solution to be found, which is then used to analyze the behavior of the tangent lines to the integral curves at given points, playing a central role in understanding the underlying geometry of differential equations.