Question

solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value \(y_{0}\). $$ y^{\prime}+y^{3}=0, \quad y(0)=y_{0} $$

Short answer

Question: Determine the particular solution and the interval of existence for the initial-value problem given by the first-order non-linear ordinary differential equation \(y^{\prime} = -y^3\) with \(y(0) = y_{0}\). Answer: The particular solution for the given initial-value problem is \(y = y_{0}(\pm e^{-2t})\), and the interval of existence for the solution is \((-∞, ∞)\).

Step-by-step solution

Rewrite the equation in separable form

To rewrite the equation, we will divide both sides by \(y^{3}\) and \(-y^{\prime}\): $$ \frac{-1}{y^{3}} y^{\prime} =1 $$ Now, the equation is in the form of a separable equation: $$ -y^{\prime} \frac{dy}{y^{3}} = dt $$

Integrate both sides

To find the general solution, we need to integrate both sides of the separable equation: $$ \int - \frac{1}{y^{3}} dy = \int dt. $$ On the left side, substitute \(u = y^{2}\), so \(du = 2y dy\). We get: $$ \int -\frac{1}{2u} du =\int dt. $$ Now, integrate both sides: $$ -\frac{1}{2}\ln|u|+C_{1} = t + C_{2} \Rightarrow -\frac{1}{2}\ln|(y^{2})|+C_{1} = t + C_{2}. $$

Apply the initial condition

Now, apply the initial condition \(y(0) = y_{0}\) to find the constant \(C_{1}\): $$ -\frac{1}{2}\ln|(y_{0}^{2})|+C_{1} = 0 \Rightarrow C_{1} = \frac{1}{2}\ln|(y_{0}^{2})|. $$ Substitute this value back into the general solution to find the particular solution: $$ -\frac{1}{2}\ln|(y^{2})|+\frac{1}{2}\ln|(y_{0}^{2})| = t. $$

Analyze the interval of existence

To find the interval of existence, we need to examine how the solution depends on the initial value \(y_{0}\): $$ -\frac{1}{2}\ln|\left(\frac{y}{y_{0}}\right)^{2}| = t. $$ Let \(x = \frac{y}{y_{0}}\). The equation becomes: $$ -\frac{1}{2}\ln|x^{2}| = t, $$ which can be solved for \(y\) by taking the exponent of both sides: $$ x = \pm e^{-2t} \Rightarrow y = y_{0}\left(\pm e^{-2t}\right). $$ The interval of existence for the solution depends on how \(t\) changes as \(y\) goes to infinity, and is not affected by the sign of \(y_{0}\). For positive and negative \(t\), the solution approaches zero, thus the interval of existence is all real numbers: $$ (-\infty, \infty). $$

Key concepts

Separable Differential Equation

A separable differential equation is a specific kind of differential equation that can be rewritten so that all terms involving one of the variables are on one side of the equation and all terms involving the other variable are on the opposite side. This allows us to solve the equation through a process called separation of variables. To identify if a differential equation is separable, the equation should be expressible in the form:

• \( rac{dy}{dx} = g(y)h(x) \)

In our initial value problem, the differential equation is \( y' + y^3 = 0 \). By rearranging terms, we rewrite it as \( y' = -y^3 \). Subsequently, we can separate variables by dividing both sides to get \( -y' \, \frac{dy}{y^3} = dt \). Now, both sides are ready for integration.
Understanding and recognizing separable differential equations are crucial because they simplify the process of solving differential equations and lead us to a particular solution effectively.

Interval of Existence

The interval of existence refers to the range of the independent variable (typically \( x \) or \( t \)) for which a solution to the differential equation is valid and meaningful. In our exercise, after separating the variables, we found the general solution:

• \( y = y_{0}(\pm e^{-2t}) \)

Our goal is to determine over what interval this solution exists. In many cases, singularities or undefined expressions might restrict the interval.
Here, analyzing the exponential decay \( e^{-2t} \), we see that as \( t \) approaches \( \pm\infty \), the term tends to zero, not introducing any discontinuities or undefined behaviors. The solution is valid for all real \( t \), thus the interval of existence is:

• \( (-\infty, \infty) \)

The initial condition \( y(0) = y_{0} \) assures that for any given initial \( y_{0} \), this interval remains the same across the real line.

Particular Solution

A particular solution of a differential equation is a solution that not only satisfies the differential equation itself but also adheres to any initial conditions provided. In the context of our problem, the initial condition given is \( y(0) = y_{0} \). We progress from the general solution to the particular solution by applying this initial condition.

• After obtaining the integrated general form: \( -\frac{1}{2}\ln|y^2| + \frac{1}{2}\ln|y_0^2| = t \)

Substituting \( y(0) = y_{0} \) provides us a way to determine the constant of integration, helping us tailor this solution specifically for the problem at hand. As a result, we resolve \( -\frac{1}{2}\ln|(y/y_0)^2| = t \), and ultimately derive:

• \( y = y_{0}(\pm e^{-2t}) \)

This expression is the particular solution, representing the behavior of \( y(t) \) that satisfies the set conditions.

Integration

Integration is a mathematical process of finding a function that describes the sum or area underneath a curve, and it is a core part of solving differential equations. In separable differential equations, when variables have been separated, integration becomes the key tool for finding the general solution.
For the equation in this exercise, after separation, we integrated both sides:

• Left-hand Side: \( \int -\frac{1}{y^3} \, dy \)
• Right-hand Side: \( \int \, dt \)

The left side uses a substitution technique where \( u = y^2 \), transforming the integral for ease of calculation, leading to a form \( -\frac{1}{2}\ln|u| \). The right side is a straightforward integration resulting in \( t + C_2 \).
This process of integrating and then applying limits or conditions bridges us from the differential form to a specific equation describing the dependent variable's change concerning the independent variable.