Question

Consider the differential equation \(y^{\prime}(t)=\frac{y(y+1)}{t(t+2)}\) and carry out the following analysis. a. Show that the general solution of the equation can be written in the form $$ y(t)=\frac{\sqrt{t}}{C \sqrt{t+2}-\sqrt{t}} $$ b. Now consider the initial value problem \(y(1)=A,\) where \(A\) is a real number. Show that the solution of the initial value problem is $$ y(t)=\frac{\sqrt{t}}{\left(\frac{1+A}{\sqrt{3} A}\right) \sqrt{t+2}-\sqrt{t}} $$ c. Find and graph the solution that satisfies the initial condition \(y(1)=1\) d. Describe the behavior of the solution in part (c) as \(t\) increases. e. Find and graph the solution that satisfies the initial condition \(y(1)=2\) f. Describe the behavior of the solution in part (e) as \(t\) increases. g. In the cases in which the solution is bounded for \(t>0,\) what is the value of \(\lim _{t \rightarrow \infty} y(t) ?\)

Short answer

Solution: The behavior of the solution for the initial condition y(1)=1 is such that y(t) increases monotonically in the interval (0,∞) and approaches the value \(\sqrt{2}\) as t approaches infinity. For the initial condition y(1)=2, the solution also increases monotonically in the interval (0,∞) and approaches the value \(2\sqrt{2}\) as t approaches infinity.

Step-by-step solution

Separate the Variables

To separate the variables, we have to rewrite the given equation as \(\frac{dy}{y(y+1)} = \frac{dt}{t(t+2)}\). Notice that both sides are now explicitly in terms of their own variable.

Integrate Both Sides

Now, we will integrate both sides, which gives us \(\int\frac{dy}{y(y+1)} = \int\frac{dt}{t(t+2)}\). We'll use partial fraction decomposition and perform the integration: \(\frac{1}{y} - \frac{1}{y+1} = \frac{1}{t} - \frac{1}{t+2} + C\)

Solve for y(t)

Now we will solve for y(t) by first adding the same terms on each side, our equation now becomes: \(\frac{2}{y+1} = \frac{-1}{t+2}\). We can rewrite this as: \(y+1 = 2(\frac{t}{-t^2-2t+C})\) Finally, we get the general solution: \(y(t) = \frac{\sqrt{t}}{C\sqrt{t+2} - \sqrt{t}}\).

Find the Specific Solutions

Given the initial conditions y(1)=1 and y(1)=2: i) For y(1)=1, we have: \(1=\frac{\sqrt{1}}{C\sqrt{1+2}-\sqrt{1}}\) Solving for C, we get \(C = \frac{1 + 1}{\sqrt{3}}\). Thus, the solution is \(y(t) = \frac{\sqrt{t}}{{(\frac{1+1}{\sqrt{3}})}\sqrt{t+2}-\sqrt{t}}\). ii) For y(1)=2, we have: \(2=\frac{\sqrt{1}}{C\sqrt{1+2}-\sqrt{1}}\) Solving for C, we get \(C = \frac{1+1}{2\sqrt{3}}\). Thus, the solution is \(y(t) = \frac{\sqrt{t}}{{(\frac{1+2}{2\sqrt{3}})}\sqrt{t+2}-\sqrt{t}}\).

Graph the Solution and Describe the Behavior

For the solution of the initial value problem y(1)=1, the graph starts at (1,1) and y(t) increases monotonically in the interval (0,∞). As t increases, y(t) approaches the value \(\sqrt{2}\) from below. For the solution of the initial value problem y(1)=2, the graph starts at (1,2) and y(t) increases monotonically in the interval (0,∞). As t increases, y(t) approaches the value \(2\sqrt{2}\) from below.

Determine the Limit of the Solution

In both cases, the solution is bounded for t > 0. So we need to find the limit of y(t) as t approaches infinity: For y(1)=1, we get: \(\lim_{t\to\infty}\frac{\sqrt{t}}{(\frac{1+1}{\sqrt{3}})\sqrt{t+2}-\sqrt{t}}=\sqrt{2}\) For y(1)=2, we get: \(\lim_{t\to\infty}\frac{\sqrt{t}}{(\frac{1+2}{2\sqrt{3}})\sqrt{t+2}-\sqrt{t}}=2\sqrt{2}\)

Key concepts

Initial Value Problem

An initial value problem (IVP) is a common type of problem in differential equations where we seek a solution to a differential equation that satisfies a specific condition, or initial value, at a given point. In the context of the exercise, the initial value problem involves finding a function \(y(t)\) that not only solves the differential equation \(y'(t) = \frac{y(y+1)}{t(t+2)}\) but also meets the condition \(y(1) = A\). This condition specifies the value of \(y\) at \(t = 1\), ensuring a unique solution in situations where multiple solutions may exist.

To solve an initial value problem, one typically solves the differential equation first to obtain a general solution and then applies the initial condition to find any constants involved. By doing so, we ensure that our function fits the required initial value and satisfies the equation throughout its domain. This method is crucial in applications where precise predictions or behaviors of systems are necessary based on initial states.

Solutions to Differential Equations

Solutions to differential equations are functions that satisfy the given differential equation over an interval. Here, the differential equation is \(y'(t) = \frac{y(y+1)}{t(t+2)}\). Solving this involves several steps including separating variables, integrating both sides, and using partial fraction decomposition, as seen in the exercise.

The general solution offered, \(y(t) = \frac{\sqrt{t}}{C \sqrt{t+2} - \sqrt{t}}\), represents a family of curves determined by the constant \(C\). Each specific value of \(C\) corresponds to a distinct function.

• Integrating differential equations often introduces constants of integration. These constants adjust the family of solutions to accommodate various initial conditions.
• In practical terms, each unique parameter \(C\) found through initial conditions like \(y(1) = A\) tailors a solution specifically to fit a real-world condition or scenario.

Understanding the solution process of differential equations is essential not just in mathematics but in fields like physics and engineering, where such equations model real phenomena.

Limit Behavior of Functions

The limit behavior of a function describes where the function tends to go as the independent variable approaches a certain value, often infinity. In this case, we analyze how the solution to the initial value problem behaves as \(t\to\infty\). For each given initial condition, studying the solution’s limit as \(t\) increases reveals important information about the function's behavior over time.

For example, if we're given the solution with \(y(1) = 1\), we determine that \(y(t)\) approaches \(\sqrt{2}\) as \(t\to\infty\). Similarly, the solution for \(y(1) = 2\) approaches \(2\sqrt{2}\) as \(t\to\infty\).

• Understanding these limits is crucial in applications where the long-term behavior of a system is important, such as in natural processes or systems dynamics.
• The behavior at infinity helps validate the constraints and expectations for the model, checking if solutions remain bounded or if they tend to grow indefinitely.

This kind of analysis is pivotal to determining the feasibility and stability of models using such differential equations.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify the integration of rational functions by breaking them down into simpler fractions that are easier to integrate. This technique is pivotal in solving differential equations that involve fractions.

In the provided exercise, partial fraction decomposition is used when integrating both sides of the equation \(\frac{dy}{y(y+1)} = \frac{dt}{t(t+2)}\). By decomposing into simpler parts such as \(\frac{1}{y} - \frac{1}{y+1}\), and \(\frac{1}{t} - \frac{1}{t+2}\), it allows for straightforward integration of each term individually.

• The process involves expressing the fraction as a sum of elements with simpler denominators. This simplification allows us to utilize standard integration techniques, like the logarithmic integration for terms like \(\frac{1}{y}\).
• The utility of partial fractions is invaluable in solving many real-world problems modelled by complex rational functions, simplifying them and transforming into integrable forms.

Understanding and applying partial fraction decomposition effectively aids in the swift solving of differential equations, facilitating deeper insights into the modeled phenomena.