Question

(a) Determine and sketch the isoclines \(f(t, y)=c\) for \(c=-1,0\), and 1 . (b) On each of the isoclines drawn in part (a), add representative direction field filaments. $$ y^{\prime}=-y+t $$

Short answer

Question: Determine and sketch the isoclines for the given ordinary differential equation \(y^{\prime}=-y+t\) for different values of \(c\), and add representative direction field filaments on each of the isoclines. Answer: The isoclines for the given ordinary differential equation are straight lines with equations \(t-y=1\) for \(c=-1\), \(t-y=0\) for \(c=0\), and \(t-y=-1\) for \(c=1\). The direction field filaments have slopes equal to the corresponding values of \(c\); that is, a slope of \(-1\) for the isocline \(t-y=1\), a slope of \(0\) for the isocline \(t-y=0\), and a slope of \(1\) for the isocline \(t-y=-1\).

Step-by-step solution

Determine the isoclines for different values of c

First, we need to find the equations of the isoclines for \(c=-1, 0,\) and \(1\). An isocline is defined as \(y^{\prime}= -y + t = c\). So for each value of c, we need to solve the equation for the function \(y\). For \(c = -1\): \(-y + t = -1 \Rightarrow t - y = 1\) For \(c = 0\): \(-y + t = 0 \Rightarrow t - y = 0\) For \(c = 1\): \(-y + t = 1 \Rightarrow t - y = -1\)

Sketch the isoclines

Now, we will sketch the isoclines for each of the determined equations in Step 1. The general form of the isoclines is \(t-y=c\) which represents straight lines with a slope of 1 and intercepts on the \(t\) and \(y\) axes as \((c,0)\) and \((0,-c)\) respectively. 1. For \(c=-1\): The isocline is \(t - y = 1\), with intercepts at \((1, 0)\) and \((0, -1)\). Draw a straight line passing through these two points. 2. For \(c=0\): The isocline is \(t - y = 0\), with intercepts at \((0, 0)\) (origin). Draw a straight line passing through the origin and making a 45-degree angle with the \(t\)-axis. 3. For \(c=1\): The isocline is \(t - y = -1\), with intercepts at \((-1, 0)\) and \((0, 1)\). Draw a straight line passing through these two points.

Add representative direction field filaments

Now, we will add the direction field filaments on each of the isoclines drawn in Step 2. For any point \((t, y)\) on the isocline, the slope of the tangent to the solution curve (direction field filament) is the same and equals \(c\). 1. For \(c=-1\): On the isocline \(t - y = 1\), the slope of the tangent to solution curve is equal to \(-1\). Draw short line segments with a slope of \(-1\) along the isocline. 2. For \(c=0\): On the isocline \(t - y = 0\), the slope of the tangent to solution curve is equal to \(0\). Draw short horizontal line segments along the isocline. 3. For \(c=1\): On the isocline \(t - y = -1\), the slope of the tangent to solution curve is equal to \(1\). Draw short line segments with a slope of \(1\) along the isocline. This completes the sketching of isoclines and adding direction fields filaments for the given ordinary differential equation.

Key concepts

Isoclines

An isocline in a differential equation is a very useful concept. It helps to visualize where certain slopes of solution curves appear on a graph. Specifically, an isocline is a curve (a straight line in our case) on which every solution has the same slope. The key to understanding isoclines is grasping that they represent sets of points where the slope of the solution curves is constant.
To find isoclines, we typically take our differential equation in the form of \(y' = f(t, y)\) and set it equal to a constant \(c\), obtaining an equation like \(f(t, y) = c\).
For the given differential equation, \(y' = -y + t\), isoclines are found by solving \(-y + t = c\). This gives you a family of lines: \(t - y = c\), where 'c' can be any constant. The straight lines represent the isoclines, and for each of them:

• If \(c = -1\), the isocline is \(t - y = 1\) with intercepts at \((1, 0)\) and \((0, -1)\).
• If \(c = 0\), the isocline is \(t - y = 0\), simply the line \(t = y\) passing through the origin.
• If \(c = 1\), the isocline is \(t - y = -1\) with intercepts at \((-1, 0)\) and \((0, 1)\).

These lines help in sketching solution curves with the same slope \(c\).

Direction Fields

Direction fields provide a graphical way to look at solutions of differential equations. Imagine a grid laid over a plane, where each point in the field represents a tiny slope. These slopes indicate the direction the solution curve will take at that point.
They are very helpful for visualizing the behavior of solutions without needing to find the exact form of the solution itself. By plotting tiny line segments with slopes given by the differential equation at various points \((t, y)\), you create a picture showing what solution curves might look like.
In our context, given an isocline for a value of \(c\), all along the isocline the slope of the direction field will be \(c\). Therefore, on:

• The line \(t - y = 1\) (where \(c = -1\)), draw line segments with slope \(-1\).
• The line \(t = y\) (where \(c = 0\)), add horizontal line segments since the slope is \(0\).
• The line \(t - y = -1\) (where \(c = 1\)), draw segments with slope \(1\).

This use of direction fields helps in sketching approximate solutions and understanding their behavior over a range of initial conditions.

Slope Fields

Slope fields are closely related to direction fields and are sometimes used interchangeably, though a subtle distinction exists. They emphasize the actual slope of the solution at any point in the \((t, y)\) plane, which is valuable in understanding the overall behavior of differential equations.
They consist of small line segments or arrows drawn at various points \((t, y)\), where each segment's direction corresponds to the slope given by the differential equation \(y' = \).
In our example with \(y' = -y + t\), each slope can differ based on the location in the \((t, y)\) plane. By calculating the slope \(y'\) for selected points, you sketch a slope field where:

• On the line \(t - y = 1\), every segment has a slope of \(-1\).
• On the line \(t = y\), every segment is horizontal, indicating a slope of \(0\).
• On the line \(t - y = -1\), every segment shows a slope of \(1\).

These fields provide intuitive insights into the nature of solutions. They're like a map, showing possible directions the solution curves can take, with each line guiding you along the path of the solution.