Question

Find all solutions of \((x-2)(x-3) y^{\prime}+2 y\) \(=(\mathrm{x}-1)(\mathrm{x}-2)\) on each of the following intervals: (a) \((-\infty, 2)\) (b) \((2,3)\) (c) \((3, \infty)\). Prove that all solutions tend to a finite limit as \(\mathrm{x} \rightarrow 2\), but that none has a finitie limit as \(\mathrm{x} \rightarrow 3\).

Short answer

Question: Determine the behavior of the solutions of the given first-order linear differential equation \((x-2)(x-3) y'+2y =(x-1)(x-2)\) on the intervals (a) \((-\infty, 2)\), (b) \((2,3)\), and (c) \((3, \infty)\). Answer: (a) All solutions tend to a finite limit as \(x\to2\) in the interval \((-\infty, 2)\). (b) All solutions tend to a finite limit as \(x\to2\) in the interval \((2,3)\). (c) None of the solutions have a finite limit as \(x\to3\) in the interval \((3, \infty)\).

Step-by-step solution

Put the equation in standard linear form

Divide both sides of the given equation by \((x-2)(x-3)\) to obtain the first-order linear differential equation in standard form: \(y'+\dfrac{2}{x-3}y=\dfrac{x-1}{x-3}\).

Find the integrating factor

In this step, we will find the integrating factor \(I(x)\). The integrating factor can be computed as: \(I(x) = e^{\int\dfrac{2}{x-3}dx} = e^{2\int\dfrac{1}{x-3}dx}=\left|e^{2\ln|x-3|}\right| = |(x-3)^2|\).

Multiply the equation by the integrating factor

Multiply both sides of the standard linear equation obtained in step 1 by the integrating factor obtained in step 2: \(|(x-3)^2(y'+\dfrac{2}{x-3}y)| = |(x-3)^2\dfrac{x-1}{x-3}|\).

Simplify the equation and integrate

Simplify both sides of the equation and integrate to find \(y\): \(\int (x-3)^2\left(y'+\dfrac{2}{x-3} y\right) dx = \int (x-3)^2 \dfrac{x-1}{x-3} dx\). Note that the left side of the equation will be the derivative of \((x - 3)^2 y\), so we have: \((x-3)^2 y = \int (x-3)(x-1) dx + C\).

Solve for y

Solve the equation for \(y(x)\) and simplify: \(y(x) = \dfrac{\int (x-3)(x-1) dx + C}{(x-3)^2}\). Now let's analyze the behavior of the solutions in each of the given intervals. (a) For interval \((-\infty, 2)\): As \(x\to2\), the denominator is tending to zero and the numerator will tend to a finite limit. Therefore, all solutions in this interval tend to a finite limit as \(x\to2\). (b) For interval \((2,3)\): For this interval, the same argument can be made as in the previous interval. As \(x\to2\), the denominator is also tending to zero and the numerator will tend to a finite limit. Thus, all solutions in this interval tend to a finite limit as \(x\to2\). (c) For interval \((3, \infty)\): As \(x\to3\), the denominator is tending to zero. To analyze the behavior of \(y\) in this interval, we should consider the limit of the numerator. Since the numerator is not tending to a finite limit, none of the solutions has a finite limit as \(x\to3\).

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique to solve first-order linear differential equations of the form \(y' + p(x)y = q(x)\). This method involves multiplying each term of the differential equation by an integrating factor that is traditionally denoted \(I(x)\), which is chosen to make the left side of the equation conveniently become the derivative of \(y\) times \(I(x)\).

The key step is finding this \(I(x)\), which is derived by exponentiating the integral of \(p(x)\). For the equation \(y'+\frac{2}{x-3}y = \frac{x-1}{x-3}\), the integrating factor would be \(I(x) = |(x-3)^2|\), as found by integrating the coefficient \(\frac{2}{x-3}\) of \(y\).

Once the integrating factor is determined, the equation is multiplied by this factor, leading to a form where the left-hand side is the derivative of the product of \(y\) and \(I(x)\). Integrating both sides of this new equation then yields \(y\) in terms of \(x\) and a constant of integration \(C\). This method is particularly useful as it provides a systematic approach to solving an entire class of differential equations commonly encountered in advanced mathematics and physics.

Differential Equation Solutions

In the realm of calculus, solving differential equation solutions involves finding a function \(y(x)\) that satisfies a given relationship between \(y\), its derivatives, and \(x\). The solution to a first-order linear differential equation, once an integrating factor has been applied and the equation simplified, comes down to integrating the modified right-hand side and dividing by the integrating factor if necessary.

In the exercise, after applying the integrating factor, we need to integrate \(\int(x-3)(x-1)dx\) to find \(y\). The resulting integral and its simplified version directly lead to a general solution that includes a constant of integration \(C\), which can be determined by initial conditions if provided. The general solution includes all particular solutions of the differential equation, and it is essential for students to learn how to derive and interpret these solutions, especially when examining their behavior across different intervals, as in part (a), (b), and (c) of the provided exercise.

Limit Behavior of Solutions

Understanding the limit behavior of solutions to differential equations is crucial, especially when studying their behavior near critical points. The exercise provided examines the limits as \(x\) approaches specific values, here, 2 and 3.

For the interval \( (-\infty, 2) \), as \(x\) approaches 2, the denominator of our solution approaches zero while the numerator tends to a finite limit. This implies that the solution approaches a finite value, an important concept in the consideration of boundary behavior of solutions.

Conversely, for the interval \( (3, \infty)\), as \(x\) approaches 3, the numerator does not tend toward a finite limit, indicating that the solution will not approach a finite limit either. This examination of limit behavior is integral to understanding the continuity and potential singularities of solutions to differential equations.

Calculus for Competitive Exams

When preparing for calculus for competitive exams, it's essential to master techniques like the integrating factor method and understand differential equation solutions and their behaviors. Competitive exams often test not only the procedural execution of solving equations but also conceptual understanding and application to different scenarios.

Students should focus on practicing a variety of problems, including those that require examining the behavior of solutions as they approach particular points. Being able to explain and justify each step of the solution process, as seen in parts (a), (b), and (c) of the exercise, strengthens conceptual understanding and prepares students for the varied and complex problems they might encounter in exams that range from Advanced Placement (AP) Calculus to various university entrance exams and math competitions.