Question

Find the solution of each of the initial value problems in Exercises 5-8, then craft a function M-file for the solution. After insuring that your function M-file is "array smart," use the function to (i) plot the solution of the initial value problem on the given interval, and (ii) evaluate the solution at the right endpoint of the given interval. \(y^{\prime}=(1-t) y\), with \(y(0)=1\), on \([0,4]\).

Short answer

Solution: \( y(t) = e^{-\frac{1}{2}t^2 + t} \), evaluate at \( t=4 \) as \( e^{-4} \).

Step-by-step solution

Identify the Differential Equation

We are given the differential equation \( y' = (1-t) y \) with the initial condition \( y(0) = 1 \). The problem is defined on the interval \([0, 4]\).

Solve the Differential Equation

The differential equation \( y' = (1-t) y \) is a first-order linear differential equation of the form \( y' + p(t)y = 0 \). Here, \( p(t) = -(1-t) \). The integrating factor is obtained by \( e^{\int -(1-t) \, dt} = e^{-t + \frac{t^2}{2}} \). Integrating with respect to \( t \), we solve using separation of variables or an integrating factor to obtain the solution: \[ y(t) = Ce^{-\frac{1}{2}t^2 + t} \].

Apply Initial Condition

Apply the initial condition \( y(0) = 1 \) to find \( C \).\[ 1 = Ce^{0} = C \]Thus, \( C = 1 \). So, \[ y(t) = e^{-\frac{1}{2}t^2 + t} \].

Create the Function M-file

Create a function in MATLAB, say \( \texttt{ySolution(t)} \):```matlabySolution = @(t) exp(-0.5*t.^2 + t);```This makes \( \texttt{ySolution} \) an array smart function that can take vectors \( t \) and apply the solution formula element-wise.

Plot the Solution on [0,4]

Plot the solution on the interval \([0, 4]\) using MATLAB:```matlabt = linspace(0, 4, 100);plot(t, ySolution(t));title('Solution of Differential Equation');xlabel('t'); ylabel('y(t)');```This commands plots the solution over the interval.

Evaluate the Solution at the Right Endpoint

Evaluate \( y \) at \( t = 4 \) using our function:```matlaby_at_4 = ySolution(4);```The value is \( y(4) \). The result is approximately \( y(4) \approx e^{-8+4} = e^{-4} \).

Key concepts

First-Order Linear Differential Equations

First-order linear differential equations are a type of ordinary differential equation (ODE) where the highest derivative is the first derivative. They typically have a structured form, given by:\[ y' + p(t)y = g(t) \]In the equation, \(y'\) represents the derivative of \(y\) with respect to \(t\), \(p(t)\) is a function of \(t\), and \(g(t)\) is another function. The aim is to find the unknown function \(y(t)\) that satisfies this equation.
For the equation from our problem, \(y' = (1-t)y\), it can be rewritten as \(y' + (t-1)y = 0\), where \(p(t) = t-1\).
First-order linear differential equations are crucial in modeling various real-world systems in physics, engineering, and biology.

MATLAB Programming

MATLAB is a powerful programming platform designed for engineers and scientists. It is widely used for numerical computation, data analysis, and visualization. When solving differential equations, MATLAB can perform symbolic computations, create function files, and plot solutions on specified intervals.
In the context of our differential equation, we wrote a MATLAB function M-file, \( \texttt{ySolution(t)} \), using anonymous functions in MATLAB:```matlabySolution = @(t) exp(-0.5*t.^2 + t);```This syntax ensures that our function is **array smart**, meaning it can handle vectors as inputs and perform operations on each element without requiring explicit loops. This property makes MATLAB highly efficient for numerical simulations and graphical displays.

Integrating Factor Method

The integrating factor method is a key technique for solving first-order linear differential equations. It simplifies the equation so that it can be integrated easily. An integrating factor, usually represented by \( \mu(t) \), is derived from the equation itself. It is expressed as:\[ \mu(t) = e^{\int p(t) dt} \]For our problem, where \( p(t) = -(1-t) \), the integrating factor becomes:\[ \mu(t) = e^{-t + \frac{t^2}{2}} \]Multiplying through the differential equation by this integrating factor converts it into a form that can be integrated directly. The process helps us find the general solution \( y(t) = Ce^{-\frac{1}{2}t^2 + t} \), where \(C\) is a constant determined by initial conditions.

Initial Value Problems

An initial value problem (IVP) involves finding a solution to a differential equation that satisfies specific conditions at a given point. Such problems are critical in determining a unique solution out of potentially infinite possibilities.
In our exercise, the initial condition is \( y(0) = 1 \). By substituting this into the general solution, we find:\[ 1 = Ce^{0} \]Resolving this gives us \( C = 1 \). Thus, the particular solution for our initial value problem remains:\[ y(t) = e^{-\frac{1}{2}t^2 + t} \]This approach ensures the solution adheres to the behavior defined by the initial condition, offering great importance to fields such as physics and engineering, where system states at initial times dictate future behavior.