Question

Consider the differential equation $$ \frac{d y}{d x}=\frac{x}{y}-\frac{x}{1+y} $$ a. Find the 1-parameter family of solutions (general solution) of this equation. b. Find the solution of this equation satisfying the initial condition \(y(0)=1 .\) Is this a member of the 1 -parameter family?

Short answer

The general solution is given by \[ \frac{1}{2} (x-y)^2 = x - \frac{1}{2} y^2 - \frac{1}{3} y^3 + C \]. The specific solution that fulfills the initial condition \(y(0)=1\) is \[ \frac{1}{2} (x-y)^2 = x - \frac{1}{2} y^2 - \frac{1}{3} y^3 + \frac{1}{6} \], which is part of the 1-parameter family of solutions as C here is \(\frac{1}{6}\).

Step-by-step solution

Simplify the differential equation

To simplify the given differential equation, we can combine the fractions on the right side of the equation: \[ \frac{d y}{d x}=\frac{x}{y}-\frac{x}{1+y} \] Simplifying this gives: \[ \frac{d y}{d x}=\frac{x-y}{y(1+y)} \]. Then we separate the variables to get: \[ y(1+y) dy = x-y dx\]

Find the general solution

The goal now is to integrate both sides with respect to x. A substitution can be made here to make the integral simpler. Through substitution, using \(u = x-y\), \(du = dx - dy\), we can get the integral equation: \[ \int u du = \int dx - y(1+y) dy \], Integrating both sides of this equation gives: \[ \frac{1}{2} u^2 = x -\frac{1}{2}y^2 -\frac{1}{3} y^3 + C \]. Unsubstituting \(u = x - y\), we get the general solution as: \[ \frac{1}{2} (x-y)^2 = x - \frac{1}{2} y^2 - \frac{1}{3} y^3 + C \]. This is a 1-parameter (C) family of solutions.

Find the specific solution where \(y(0)=1\)

Substituting \(y=1\) and \(x=0\) into the general solution, we can find the specific constant value for this initial condition: \[ \frac{1}{2} (0-1)^2 = 0 - \frac{1}{2} (1)^2 - \frac{1}{3} (1)^3 + C \]. Solving for C from this yields \(C = \frac{1}{6}\). Substituting \(C = \frac{1}{6}\) in the general solution gives our specific solution: \[ \frac{1}{2} (x-y)^2 = x - \frac{1}{2} y^2 - \frac{1}{3} y^3 + \frac{1}{6} \]. This should satisfy the given initial condition and also be a member the 1-parameter family of the general solution, since the particular value of C could be covered by the 1-parameter family.

Key concepts

Separable Differential Equations

When confronted with a differential equation, one of the first strategies to consider is whether the equation is separable. A separable differential equation is one in which the variables can be separated on different sides of the equation. Specifically, such an equation can be written in the form \( \frac{dy}{dx} = g(x) h(y) \) where \( g(x) \) and \( h(y) \) are functions of \( x \) and \( y \), respectively. This separation allows us to rearrange the equation to have all instances of \( y \) and its differential \( dy \) on one side and all instances of \( x \) and \( dx \) on the other, as shown in the original exercise:

\[ y(1+y) dy = x-y dx \]
To solve a separable equation, we integrate both sides, often needing substitution or other algebraic manipulations to reach a solvable integral. In our example, the integration process involves a clever substitution that simplifies the integral and ultimately leads to finding the general solution.

Integrating Factors

An integrating factor is a technique used to solve certain differential equations that are not readily separable or exact. Essentially, an integrating factor is a function, often denoted by \( \mu(x) \) or \( \mu(y) \), which when multiplied by the differential equation, transforms it into an exact differential equation — one that can be integrated directly.

This method is particularly beneficial when dealing with linear first-order equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). By multiplying an appropriately chosen \( \mu(x) \) through the equation, it takes on a form that allows for the direct integration of both sides. Although an integrating factor is not required for the exercise provided, understanding this concept is invaluable for tackling a broad array of more complex differential equations where separation of variables is not possible.

Initial Value Problem

An initial value problem (IVP) is a class of differential equations accompanied by a set of initial conditions. The initial conditions specify the values of the dependent variable (and possibly its derivatives) at a certain point. For a first-order differential equation, the IVP can be written as \( \frac{dy}{dx} = f(x,y), \quad y(x_0) = y_0 \), where \( (x_0, y_0) \) defines the initial condition.

The importance of the initial value problem lies in its application to real-world scenarios where not only the pattern of change is known (the differential equation itself), but also the specific state at a known point in time (the initial condition). Solving an IVP leads to a unique solution that fits both the differential equation and the initial condition, offering a predictive power that broad differential solution families do not. As seen in the exercise, after finding a general solution, we use the initial condition \( y(0)=1 \) to identify the specific solution that traces its path through the given point.