Question

Solving Differential Equations Let \(\frac{d y}{d x}=x-\frac{1}{x^{2}}\) (a) Find a solution to the differential equation in the interval \((0,)\) that satisties \(y(1)=2\) (b) Find a solution to the differential equation in the interval \((-\infty, 0)\) that satisfies \(y(-1)=1\) (c) Show that the following piecewise function is a solution to the differential equation for any values of \(C_{1}\) and \(C_{2}\) . \(y=\left\\{\begin{array}{l}{\frac{1}{x}+\frac{x^{2}}{2}+C_{1}} \\\ {\frac{1}{x}+\frac{x^{2}}{2}+C_{2}}\end{array}\right.$$x<0\) \(x>0\) (d) Choose values for \(C_{1}\) and \(C_{2}\) so that the solution in part (c) agrees with the solutions in parts (a) and (b). (e) Choose values for \(C_{1}\) and \(C_{2}\) so that the solution in part (c) satisfies \(y(2)=-1\) and \(y(-2)=2\)

Short answer

The respective specific solutions for part (a) and (b) are both \(y(x)=\frac{x^2}{2} + \frac{1}{x} + 1\). For part (c), for any \(C_{1}\) and \(C_{2}\), the piecewise function indeed satisfies the given differential equation. For parts (d) and (e), the values for \(C_{1}\) and \(C_{2}\) are \(1\) and \(-\frac{5}{2}\), \(\frac{7}{2}\) respectively.

Step-by-step solution

Solving Given Differential Equation

The given first order differential equation is \(\frac{d y}{d x}=x-\frac{1}{x^{2}}\). To solve it, calculation leads to its general solution as \(y(x)=\frac{x^2}{2} + \frac{1}{x} + C\). This will be used to find the specific solutions in the following steps.

Find Solution for Interval (0,∞) with Condition y(1)=2

Substitute the initial condition \(y(1)=2\) into the general solution, which leads to \(2=\frac{1^2}{2} + \frac{1}{1} + C\), solving gives \(C=1\). So the solution is \(y(x)=\frac{x^2}{2} + \frac{1}{x} + 1\) for interval (0,∞).

Find Solution for Interval (-∞,0) with Condition y(-1)=1

Similarly, substitute the initial condition \(y(-1)=1\) into the general solution as \(1=\frac{(-1)^2}{2} - \frac{1}{-1} + C\), solving gives \(C=1\). The solution is \(y(x)=\frac{x^2}{2} + \frac{1}{x} + 1\) for interval (-∞,0).

Verifying the Piecewise Function as a Solution

For any values of \(C_{1}\) and \(C_{2}\), differentiation of the given piecewise function gives \(\frac{d y}{d x}=x-\frac{1}{x^{2}}\), meaning it is indeed a solution to this differential equation.

Determine \(C_{1}\) and \(C_{2}\) for Solutions in Part (a) and (b)

Since the solutions in parts (a) and (b) are both \(y(x)=\frac{x^2}{2} + \frac{1}{x} + 1\), for the function to agree with these solutions, it requires \(C_{1}=C_{2}=1\).

Determine \(C_{1}\) and \(C_{2}\) for \(y(2)=-1\) and \(y(-2)=2\)

Substitute \(x=2\) and \(x=-2\) into the function respectively, the equations \(C_{1}=-\frac{5}{2}\) and \(C_{2}=\frac{7}{2}\) are obtained. Thus, \(C_{1}=-\frac{5}{2}\) and \(C_{2}=\frac{7}{2}\) are the values that satisfy the given conditions.

Key concepts

Initial Value Problem

An initial value problem (IVP) arises when you are given a differential equation and you are asked to find a function that not only satisfies the differential equation, but also meets a specified condition at a certain point, often called the initial condition. This is like being handed a puzzle where the pieces are scattered, and one crucial piece is already in place to help guide the rest of the construction.

In the context of our exercise, the differential equations provided come with conditions like `y(1)=2` and `y(-1)=1`. These conditions act as the starting points from which we calculate the constant value `C` after integrating the differential equation. Solving an initial value problem thus involves two main steps: first, finding the general solution of the differential equation through integration, and second, applying the initial condition to pinpoint the exact solution specific to the problem at hand.

General Solution of Differential Equation

The general solution of a differential equation contains all possible solutions to the equation and includes a constant or a function that represents an arbitrary element. Think of the general solution as a template: it gives a comprehensive form of the solution without being bound to any particular initial condition. It’s a family photo that includes everyone, even those cousins whose names you might not remember.

When we talk about solving a differential equation, we are typically looking for this general solution. It may include constants of integration, which represent the infinite set of possible specific solutions. For example, in our problem, the general solution derived from integrating the given differential equation is `y(x)=x^2/2 + 1/x + C`. Notice the `C` here, which will change depending on the initial conditions provided.

Piecewise Function

A piecewise function is essentially a mathematical chameleon, taking on different forms or rules for different intervals of its domain. It's as if you have a playlist with different tunes to match your changing moods throughout the day.

In Step 4 of the solution steps, we encounter a piecewise function where `y(x)` is defined by two different expressions depending on whether `x` is greater than or less than zero. This allows the function to behave differently on either side of the x-axis, providing a tailored solution to the differential equation that can still meet multiple conditions across its domain. The seamless transition between these ‘pieces’ of the function is vital and often requires that we choose constants wisely to ensure the function is continuous, or at least well-behaved, at the points where the expression changes.

Integration

Integration is the mathematical tool we use to solve differential equations—like finding the source of a river by tracing its flow upstream. It allows us to reconstruct the original function from its rate of change or derivative. When we integrate a function, we are looking for the antiderivative, which is a function whose derivative is the function we started with.

In our problem, we must integrate the function `dy/dx=x-1/x^2` to find `y`, the function that describes the relationship between `x` and `y` before differentiation occurred. This process not only establishes the general solution but also lays the groundwork for finding specific solutions by applying initial conditions. By integrating and then applying the initial values, we move from the broad possibilities of the general solution to the precision of a particular solution that adheres to the conditions we're given.