Question

Consider the initial value problem $$ y^{\prime}=t y(4-y) / 3, \quad y(0)=y_{0} $$ $$ \begin{array}{l}{\text { (a) Determine how the behavior of the solution as } t \text { increases depends on the initial }} \\ {\text { value } y_{0} \text { . }} \\ {\text { (b) Suppose that } y_{0}=0.5 \text { . Find the time } T \text { at which the solution first reaches the value }} \\ {3.98 \text { . }}\end{array} $$

Short answer

In this problem, we are given a first-order nonlinear differential equation, and our goal is to analyze the behavior of the solution as time increases based on the initial value (part a) and determine the time at which the solution first reaches a specified value (part b). After solving the given differential equation, we found the general solution: $$ y = \frac{4Ke^{\frac{t^2}{6}}}{1 + Ke^{\frac{t^2}{6}}} $$ 1. In part (a), we applied the initial condition and analyzed the behavior of the solution as time increased: - If $04$, the initial values are not compatible with the given differential equation. 2. In part (b), we found the time $T$ at which the solution first reaches the value 3.98 with an initial value of $y_0 = 0.5$. We determined that $T \approx 4.01$.

Step-by-step solution

Rewrite and separate the variables

First, rewrite the given differential equation as a separable equation by dividing both sides by \(ty(4-y)\) and rearranging the terms: $$ \frac{dy}{y(4-y)} = \frac{t}{3} dt $$ Now we have separated the variables.

Integrate both sides

To find the general solution of the differential equation, integrate both sides of the equation: $$ \int \frac{1}{y(4-y)} dy = \int \frac{t}{3} dt $$ The left side can be integrated using partial fractions: $$ \frac{1}{y(4-y)} = \frac{A}{y} + \frac{B}{4-y} $$ Solving for A and B gives: $$ A = 1, \quad B = -1 $$ Thus, our integral becomes: $$ \int \left( \frac{1}{y} - \frac{1}{4-y} \right) dy = \int \frac{t}{3} dt $$ Now, we can integrate both sides: $$ (\ln |y| - \ln |4-y|) = \frac{t^2}{6} + C $$

Solve for y

Now, we solve for y: $$ \ln \left| \frac{y}{4-y} \right| = \frac{t^2}{6} + C $$ Remove the logarithms: $$ \frac{y}{4-y} = e^{\frac{t^2}{6} + C} = e^{C}e^{\frac{t^2}{6}} $$ Let \(K = e^C\): $$ y = (4 - y)Ke^{\frac{t^2}{6}} $$ Next, we solve for \(y\) and simplify: $$ y = \frac{4Ke^{\frac{t^2}{6}}}{1 + Ke^{\frac{t^2}{6}}} $$ This is the general solution of the given differential equation.

Apply the initial condition and analyze the behavior (Part a)

Determine the value of K when \(y(0) = y_0\): $$ y_0 = \frac{4K}{1+K} $$ Now, we can analyze the behavior as \(t\) increases by looking at the limiting values of the solution. Applying the limit as \(t\) goes to infinity: $$ y(t) = \lim_{t \rightarrow \infty} \frac{4Ke^{\frac{t^2}{6}}}{1 + Ke^{\frac{t^2}{6}}} $$ Because \(\frac{t^2}{6}\) becomes a dominant term as \(t\) increases, we observe the behavior for various initial values \(y_0\): 1. If \(04\), we cannot find a value for K that satisfies the initial condition, so these initial values are not compatible with the given differential equation.

Find the time T at which the solution reaches the specified value (Part b)

We are given that the initial value \(y_0 = 0.5\). Now we need to find the time \(T\) at which the solution first reaches the value 3.98. Substitute the initial condition and the specified value into the general solution: $$ 0.5 = \frac{4K}{1+K}, \quad 3.98 = \frac{4Ke^{\frac{T^2}{6}}}{1 + Ke^{\frac{T^2}{6}}} $$ First, solve for K from the initial condition equation: $$ K = \frac{0.5}{3.5} = \frac{1}{7} $$ Substitute K into the equation for the specified value and solve for T: $$ 3.98 = \frac{4(\frac{1}{7})e^{\frac{T^2}{6}}}{1 + (\frac{1}{7})e^{\frac{T^2}{6}}} $$ We can simplify the equation, and then solve for \(T^2\): $$ \frac{279}{2} = e^{\frac{T^2}{6}} $$ $$ T^2 = 6\ln{\left(\frac{279}{2}\right)} $$ Now take the square root to find T: $$ T \approx 4.01 $$ Thus, the solution first reaches the value 3.98 at approximately \(T = 4.01\).

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In our context, the differential equation describes how a quantity changes over time when that change is proportional to the quantity's current value and possibly external factors, represented by variables such as time (t).

The initial value problem provided, involves finding a function y(t) that satisfies the differential equation y' = ty(4-y)/3, given that y has a specific initial value y_0 when t=0. This kind of problem is fundamental in various scientific and engineering fields as it describes the evolution of processes over time.

Separable Equation

A separable equation is a type of differential equation in which the variables can be separated on opposite sides of the equation, facilitating its solution through integration. In the step-by-step solution, we manipulate the original differential equation to express it in the form dy/f(y) = g(t)dt, where f(y) and g(t) are functions of y and t, respectively.

This manipulation makes the problem solvable through standard integration techniques. In our example, separating the variables allows us to integrate both sides with respect to their own variable, setting the stage for finding the explicit solution y(t).

Partial Fractions

The technique of partial fractions is a method used to integrate rational functions, which are ratios of polynomials. It involves breaking down a complex fraction into simpler ones that can be easily integrated. This method is particularly useful when dealing with the integration of fractions involving products of linear factors.

In our exercise, to integrate 1/y(4-y), we express it as the sum of two simpler fractions with unknown coefficients A and B. By finding these coefficients, we can integrate each of the resulting simpler fractions individually and then combine the results to get the integrated value of the original expression.

Integration

Integration is one of the two main operations in calculus, alongside differentiation. It can be interpreted as finding the area under the curve of a function. When solving differential equations, integration allows us to retrieve the original function from its derivative.

In the provided solution, we integrate both sides of the separable equation after applying partial fractions, resulting in the integral of dy/(y(4-y)) on the left side and the integral of t/3 on the right side. These integrals bring us closer to obtaining the function y(t) that we're seeking.

Limit Behavior

The limit behavior of a function refers to the values that the function approaches as the input either becomes extremely large or small. In terms of differential equations, it is particularly interesting to study the limit behavior of solutions as it tells us about the long-term behavior of the dynamic system described by the equation.

For the given initial value problem, it's important to discover what happens to y(t) as t increases indefinitely. Does it approach a specific value, does it continue to grow without bound, or does it oscillate? The answer depends on the initial value y_0 and the specifics of the differential equation, informing us how the system evolves over time.

Initial Condition

The initial condition of a differential equation specifies the starting state, or the known value of the function at a particular point, often at t=0. Initial conditions are crucial because, without them, a unique solution to a differential equation cannot be guaranteed.

In our exercise, the initial condition y(0) = y_0 enables us to find the specific constant of integration, represented as K in the solution after integrating. This constant is determined by the initial value of y and ensures that the solution we find to the differential equation is the one that precisely fits the initial value problem presented.