Question

Slope Field In Exercises \(67-72,\) use a computer algebra system to (a) graph the slope field for the differential equation and (b) graph the solution satisfying the specified initial condition. $$ \frac{d y}{d x}=0.4 y(3-x), \quad y(0)=1 $$

Short answer

The solution is \(y = e^{0.4(3x - 0.5x^2)}\). This solution curve passes through the point (0,1), satisfying the given initial condition.

Step-by-step solution

Plot the Slope Field

A slope field is a way of visualizing a family of solutions to a first order differential equation by plotting arrows representing the slope at different points. Insert the given first order differential equation, \(\frac{d y}{d x}=0.4 y(3-x)\), into a computer algebra system (such as Matlab, Mathematica, or SageMath) to plot the slope field. This provides a visual approximation of possible solutions to the differential equation.

Solve the differential equation

Integrate the given differential equation, \(\frac{d y}{d x}=0.4 y(3-x)\). One can do this by separating variables and applying the integral on both sides of the equation. Thus, it becomes \(\int \frac{1}{y}\, dy = 0.4 \int (3-x)\, dx\). Solving these integrals gives \(\ln|y| = 0.4(3x - 0.5x^2) + C\) where C is the constant of integration. Raising e to both sides gives \(y = e^{0.4(3x - 0.5x^2) + C} = A e^{0.4(3x - 0.5x^2)}\) where \(A = e^C\) is an arbitrary constant.

Apply the initial conditions

Apply the given initial condition \(y(0)=1\). Plug in \(0\) for \(x\) and \(1\) for \(y\) into the solution \(y = A e^{0.4(3x - 0.5x^2)}\). Solving for A, we get \(A = 1\).

Graph the solution curve

The general solution is the family of solutions to the differential equation. The specific solution is the curve that passes through the given point (0,1). Graph the specific solution \(y = e^{0.4(3x - 0.5x^2)}\) on the same coordinate plane as the slope field.

Key concepts

First Order Differential Equation

A first order differential equation is a relation that involves rates of change and the function itself. In our exercise, the equation \( \frac{d y}{d x}=0.4 y(3-x) \) is a first order because it involves the first derivative of the function y with respect to x. Such equations are fundamental in expressing how a quantity changes over time or another variable.

Understanding first order equations is crucial because they often model real-world scenarios such as population growth, chemical reactions, and body temperature change. They can vary greatly in complexity, and the slope field graph is an excellent tool for visualizing their solutions, depicting how the slope varies depending on the location in the plane.

Separation of Variables

The technique of separation of variables allows for the isolation of different variables on opposite sides of an equation, which can then be integrated separately. It is a commonly used method for solving first order differential equations. In our example, the equation can be rearranged so that all terms involving y are on one side, and all terms involving x are on the other, leading to \( \int \frac{1}{y}\, dy = 0.4 \int (3-x)\, dx \).

Once variables are separated, each side can be integrated independently, allowing us to derive a general solution and later, apply initial conditions to find a specific solution. This method greatly simplifies the process of solving many differential equations.

Computer Algebra System

A computer algebra system (CAS) is a software tool that facilitates symbolic mathematical computations. This means it can solve equations, perform algebraic manipulations, and visualize mathematical concepts like slope fields and solution curves. Examples of CAS include MATLAB, Mathematica, and SageMath, among others.

Using a CAS allows for accurate and efficient plotting of complex graphs or solving intricate equations. For students, computer algebra systems can be invaluable in checking work done by hand, experimenting with variations of problems, and fostering a deeper understanding of the material by enabling interactive learning experiences.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with a specified value, called an initial condition, at a particular point. Solving such a problem involves finding a function that satisfies both the differential equation and the initial condition. In our exercise, the initial condition is given by \( y(0)=1 \).

This specific 'point' tells us where the solution curve should start or pass through. When solving differential equations, initial conditions allow us to narrow down the infinite set of possible solutions to the one that actually represents the scenario we’re modeling. Applying the initial condition guides us towards the particular solution that is relevant to the problem at hand.

Integrating Factor

An integrating factor is a function that is typically used to solve linear nonhomogeneous first order differential equations that are not separable. The purpose of an integrating factor is to multiply the original equation by this factor to create a left-hand side that is the derivative of a product of two functions.

However, in our specific exercise, we do not require an integrating factor because the given differential equation is directly separable. In situations where separation of variables is not possible, the integrating factor is an invaluable tool, often being a function of the independent variable x, which, after being multiplied by the differential equation, simplifies the equation to a form that can easily be integrated.