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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

Expert verified
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

01

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: $$ -y^{\prime} \frac{dy}{y^{3}} = dt $$
02

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: $$ -\frac{1}{2}\ln|u|+C_{1} = t + C_{2} \Rightarrow -\frac{1}{2}\ln|(y^{2})|+C_{1} = t + C_{2}. $$
03

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: $$ -\frac{1}{2}\ln|(y^{2})|+\frac{1}{2}\ln|(y_{0}^{2})| = t. $$
04

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: $$ (-\infty, \infty). $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A recent college graduate borrows 100,000 dollar at an interest rate of \(9 \%\)to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800 dollar (1+t / 120), where t is the number of months since the loan was made. $$ \begin{array}{l}{\text { (a) Assuming that this payment schedule can be maintained, when will the loan be fully }} \\ {\text { paid? }} \\ {\text { (b) Assuming the same payment show large a loan could be paid off in exactly }} \\ {\text { 20 years? }}\end{array} $$

Consider the initial value problem \(y^{\prime}=y^{1 / 3}, y(0)=0\) from Example 3 in the text. (a) Is there a solution that passes through the point \((1,1) ?\) If so, find it. (b) Is there a solution that passes through the point \((2,1)\) ? If so, find it. (c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at \(t=2\)

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with theoretical limiting concentration if the tank had infinite capacity.

Show that any separable equation, $$ M(x)+N(y) y^{\prime}=0 $$ is also exact.

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. In this problem we assume that fish are caught at a constant rate \(h\) independent of the size of the fish population. Then \(y\) satisfies $$ d y / d t=r(1-y / K) y-h $$ The assumption of a constant catch rate \(h\) may be reasonable when \(y\) is large, but becomes less so when \(y\) is small. (a) If \(hy_{0}>y_{1},\) then \(y \rightarrow y_{2}\) as \(t \rightarrow \infty,\) but that if \(y_{0}r K / 4,\) show that \(y\) decreases to zero as \(l\) increases regardless of the value of \(y_{0}\). (c) If \(h=r K / 4\), show that there is a single cquilibrium point \(y=K / 2\) and that this point is semistable (see Problem 7 ). Thus the maximum sustainable yield is \(h_{m}=r K / 4\) corresponding to the equilibrium value \(y=K / 2 .\) Observe that \(h_{m}\) has the same value as \(Y_{m}\) in Problem \(20(\mathrm{d})\). The fishery is considered to be overexploited if \(y\) is reduced to a level below \(K / 2\).

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