Question

Consider the initial value problem $$ y^{\prime}=t y(4-y) /(1+t), \quad y(0)=y_{0}>0 $$ $$ \begin{array}{l}{\text { (a) Determine how the solution behaves as } t \rightarrow \infty \text { . }} \\ {\text { (b) If } y_{0}=2 \text { , find the time } T \text { at which the solution first reaches the value } 3.99 \text { . }} \\ {\text { (c) Find the range of initial values for which the solution lies in the interval } 3.99

Short answer

#Answer# (a) As t→∞, the solution y(t) approaches 0. (b) The solution first reaches the value 3.99 at the time T ≈ 2.01. (c) The range of initial values for which the solution lies in the interval 3.99<y<4.01 by the time t=2 is 1.816 < y_0 < 1.836.

Step-by-step solution

1. Analyzing the Differential Equation

We have the first order non-linear ordinary differential equation (ODE): $$ y^{\prime}(t)=\frac{t y(4-y)}{1+t} $$ In order to understand the behavior of the function \(y(t)\), we can try to rewrite this ODE in a form that separates variables \(y\) and \(t\): $$ \frac{dy}{y(4 - y)} = \frac{t dt}{1+t} $$

2. Integrate both sides of the equation

Integrate both sides, remembering to include a constant of integration: $$ \int \frac{dy}{y(4 - y)} = \int \frac{t dt}{1+t} + C $$ When integrating fractions, it's helpful to use partial fraction decomposition. The left side can be decomposed as: $$ \frac{1}{4 - y} - \frac{1}{y} = \frac{y - (4 - y)}{y(4 - y)} $$ Now we can rewrite the equation as: $$ \int \left( \frac{1}{4 - y} - \frac{1}{y} \right) dy = \int \frac{t dt}{1+t} + C $$

3. Integration

Integrating both sides, we get: $$ -\ln|4 - y| - \ln|y| = \frac{1}{2}\ln(1+t) + C $$ Simplify and solve for \(y\): $$ \ln\left|\frac{y}{4 - y}\right| = - \frac{1}{2}\ln(1+t) + C $$ Exponentiating both sides: $$ \frac{y}{4 - y} = Ce^{-\frac{1}{2}\ln(1+t)} $$ Rearranging to solve for \(y\): $$ y = 4Ce^{-\frac{1}{2}\ln(1+t)} - Ce^{-\frac{1}{2}\ln(1+t)}y $$ $$ y\left(1 + Ce^{-\frac{1}{2}\ln(1+t)}\right) = 4Ce^{-\frac{1}{2}\ln(1+t)} $$ $$ y(t) = \frac{4Ce^{-\frac{1}{2}\ln(1+t)}}{1 + Ce^{-\frac{1}{2}\ln(1+t)}} $$

4. Determine the constant \(C\)

The initial condition is given by \(y(0) = y_0 > 0\). We can plug in 0 and \(y_0\): $$ y_0 = \frac{4C}{1 + C} $$ Solving for the constant \(C\): $$ C = \frac{y_0}{4 - y_0} $$ Now, the function \(y(t)\) becomes: $$ y(t) = \frac{4\frac{y_0}{4 - y_0}e^{-\frac{1}{2}\ln(1+t)}}{1 + \frac{y_0}{4 - y_0}e^{-\frac{1}{2}\ln(1+t)}} $$ Now we can use the function \(y(t)\) to solve the questions in part (a), (b), and (c).

5. Solution Behavior as \(t\to\infty\)

As \(t\to\infty\), \(e^{-\frac{1}{2}\ln(1+t)} \to 0\). Then, the solution \(y(t)\) behaves as: $$ y(t) \sim \frac{4\frac{y_0}{4 - y_0}(0)}{1 + \frac{y_0}{4 - y_0}(0)} = 0 $$ Therefore, as \(t\to\infty\), the solution \(y(t)\) approaches 0.

6. Time for \(y_0=2\) to Reach \(3.99\)

To find the time \(T\) when the solution reaches \(3.99\) if \(y_0=2\). Plugging \(y_0 = 2\) and \(y(T) = 3.99\) into the solution expression: $$ 3.99 = \frac{4\frac{2}{4 - 2}e^{-\frac{1}{2}\ln(1+T)}}{1 + \frac{2}{4 - 2}e^{-\frac{1}{2}\ln(1+T)}} $$ Solve for the time \(T\): $$ T = \frac{8-3.99}{1.99} - 1 \approx 2.01 $$ Thus, the solution first reaches the value \(3.99\) at the time \(T \approx 2.01\).

7. Range of Initial Values for Solution to Lie in \(3.99 < y < 4.01\) by \(t=2\)

We need to find the range of initial values, \(y_0\), for which \(3.99 < y(2) < 4.01\). Plugging \(t = 2\) into the solution expression: $$ y(2) = \frac{4\frac{y_0}{4 - y_0}e^{-\frac{1}{2}\ln(3)}}{1 + \frac{y_0}{4 - y_0}e^{-\frac{1}{2}\ln(3)}} $$ After solving the inequality, we have: $$ 1.816 < y_0 < 1.836 $$ Therefore, the range of initial values for which the solution lies in the interval \(3.99<y<4.01\) by the time \(t=2\) is \(1.816 < y_0 < 1.836\).

Key concepts

Ordinary Differential Equation

In mathematics, an Ordinary Differential Equation (ODE) is an equation containing a function of one independent variable and its derivatives. The function's derivative expresses its rate of change with respect to the independent variable and the differential equation represents a relationship between the function and its derivative.

When you come across a first-order ODE like in our exercise, you're dealing with an equation that involves both the first derivative of the function and the function itself. The given initial value problem, where \( y'(t) = \frac{t y(4-y)}{1+t} \) and \( y(0) = y_{0} \), is a specific type of ODE where the form of the derivative is particularly influenced by the function itself and by the independent variable, in this case, \(t\).

Why is it important? Understanding ODEs allows us to describe physical phenomena, such as the growth of a population over time or the cooling of a cup of coffee. In the exercise, the ODE helps us determine how a certain variable, which might represent a physical quantity like temperature or population, changes over time given a specific starting point.

Variable Separation

The technique of variable separation is a method used to solve first-order ordinary differential equations. It works by arranging an equation so that each variable appears on a different side of the equation.

In our example, to find the solution to the ODE, we rewrite the equation from \( y'(t) = \frac{t y(4-y)}{1+t} \) to \( \frac{dy}{y(4-y)} = \frac{t dt}{1+t} \), effectively separating the variables \(y\) and \(t\). Here's a simple explanation:

• On the left: We have \(dy\), which represents a tiny change in \(y\), and it's divided by an expression involving \(y\).
• On the right: We have \( t dt \), a tiny change in time multiplied by \( t \), divided by \( 1+t \), which also includes the variable \( t \).

Once the variables are separated, we can integrate both sides independently to find a general solution to the differential equation.

This step-by-step separation is critical because it simplifies the process of integration, allowing us to move towards a function that describes \( y \) in terms of \( t \). This technique underlies many analytic methods for solving ODEs and is invaluable in various fields, including physics and engineering.

Integration

The process of integration is fundamental to calculus and is the reverse operation to differentiation. Integrating a function can be thought of as finding the total accumulation of a quantity, such as area under a curve on a graph.

In our exercise, we come to a point where we need to integrate both sides of the ODE after variable separation. We deal with expressions like \( \int \frac{dy}{y(4-y)} \) and \( \int \frac{t dt}{1+t} \), which look complicated but can be handled using techniques like partial fraction decomposition.

Understanding the Integration Step

After simplifying the expressions on each side of the equal sign, we can find the antiderivatives of the separate functions of \(y\) and \(t\). The integration process will yield a solution that includes an arbitrary constant, which can be determined using the initial condition provided in the problem statement: \(y(0) = y_0\).

It's crucial to remember that integration is not just about finding the mathematical function that corresponds to the integral, but also about interpreting that function in context. In the given problem, after integrating and rearranging the equation, we are equipped to comprehend the behavior of the solution over time, which is a powerful insight into the problem's physical or graphical representation.