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In this lesson, we’ll learn how to draw slope fields, which help visualize the general solution of first-order differential equations graphically.
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We’ll determine which equation a particular slope field represents.
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And in reverse, we’ll work out which slope field represents a particular differential equation.
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But first, let’s clarify some of the terms we’re using.
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Suppose we have an unknown function 𝑦, which is a function of 𝑥.
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The slope of 𝑦 d𝑦 by d𝑥 is the derivative of 𝑦 with respect to 𝑥.
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A differential equation is an equation containing the derivative of a function.
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A first-order differential equation contains only the first derivative.
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And a differential equation defines the slope in terms of a function 𝑓.
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𝑓 can be a function of 𝑥.
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𝑓 can be a function of 𝑦.
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Or 𝑓 can be a function of 𝑥 and 𝑦.
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Ideally, given the differential equation as our starting point, we’d like to be able to solve it.
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This means find the antiderivative, that is, the function 𝑦 of 𝑥.
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But as you probably know, for more differential equations than not, this is not possible.
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There are actually very few differential equations that can be solved exactly.
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All is not lost, however, because what we can do is plot the derivative of the function.
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That’s the differential equation at various points in the 𝑥𝑦-plane.
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This is called a slope field or the direction field.
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In this example, d𝑦 by d𝑥 is equal to two 𝑥, which is a function of 𝑥.
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Each of the blue lines represents the slope of a particular solution 𝑦 of 𝑥 at that point.
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So, for example, when 𝑥 is equal to approximately 0.5, the slope of the solutions at that point is positive.
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In fact, we can find the slope for any solution for 𝑥 is 0.5 by substituting 𝑥 is 0.5, which is a half, into the differential equation.
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And we find d𝑦 by d𝑥 is two times a half, which is equal to one.
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So that at approximately 𝑥 is equal to 0.5, for any solution to this differential equation, the slope is one at that point.
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A particular solution to a differential equation is defined using an initial value.
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So, for example, in our case, we have three particular solutions on this graph, where solution A has an initial value of zero, one.
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Solution B has an initial value of zero, zero.
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And solution C has an initial value of zero, negative two.
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In fact, the general solution to this differential equation is 𝑦 of 𝑥 is equal to 𝑥 squared plus a constant of integration.
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For the particular solution A, the constant of integration is equal to one.
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For solution B, the constant of integration is equal to zero.
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And for solution C, the constant of integration is equal to negative two.
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These are just three possible solutions to this differential equation.
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There are actually an infinite number, each defined by the constant 𝐶.
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And remember that the slope field plot gives us some idea of the general behavior of the family of solutions of the differential equation.
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With an initial value, we can trace out a particular solution.
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And this solution follows the direction of the slope of the line segments in the slope field.
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Let’s look now at how we might actually draw a slope field.
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Suppose we have the differential equation d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦, which is a function of 𝑥 and 𝑦.
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What this tells us is that, at any point in the 𝑥𝑦-plane, the slope of the solution of the differential equation at that point is equal to the 𝑥-coordinate minus the 𝑦-coordinate.
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And we plot a small line with this slope at this point in the 𝑥𝑦-plane.
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Let’s do this for some example points using our differential equation.
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If we take 𝑥 is zero and 𝑦 is one as our first point, the slope d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦, which is zero minus one.
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And that’s equal to negative one.
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So the slope at zero, one is negative one.
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Let’s start putting our points on the graph so we can see how this looks in the slope field.
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At the point zero, one, our solution has a slope of negative one.
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Let’s choose as our next point 𝑥 is zero and 𝑦 is zero.
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In this case, d𝑦 by d𝑥 is equal to zero minus zero, which is equal to zero.
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So at this point, we have a horizontal line which has a slope of zero.
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Choosing one, zero as our next point, the slope is one minus zero, which is one.
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At the point one, zero therefore, we have the positive slope of one.
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If we continue in this way, the point zero, negative one has a slope of one.
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The point negative one, zero has a slope of negative one.
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The point one, one has a slope of zero.
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And the point negative one, negative one also has a slope of zero.
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If we continue in this way for many more points in the 𝑥𝑦-plane, we get the slope field.
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And this slope field represents the differential equation d𝑦 by d𝑥 is equal to 𝑥 minus 𝑦.
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The solution passing through a particular point on the slope field will follow the direction of these small line segments.
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We say that this is the solution graph with initial condition 𝑥 zero, 𝑦 zero, where in this case 𝑥 zero, 𝑦 zero is equal to negative one, negative one.
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Something to notice that a lower solution graph passing through any of the plotted line segments will have the same slope as the segment at that point.
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A solution might not follow only the plotted segments.
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It’s likely to weave between the segments.
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Remember that the plot is just the derivative at a selection of points.
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So a slope field is a way to represent an infinite number of specific solutions.
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And if we know that the one we’re looking for goes through a particular point, then following the slope lines from that point traces out that particular solution.
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Let’s look at an example.
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Consider the given slope field graph representing a differential equation.
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If the solution of the differential equation contains point 𝑆, which point can also belong to the solution?
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Let’s trace solution curves through each of the points in the direction of the pattern of slopes mapped out through the line segments.
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Starting at the point in question, 𝑆, we can see that the solution curve could go through the point 𝐶.
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Although 𝐶 doesn’t lie exactly on the line segment, it is within the pattern of the slope field marked out by a solution through the point 𝑆.
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So point 𝐶 is a contender for the solution through 𝑆.
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Let’s also check the points 𝐴, 𝐵, 𝐷, and 𝐸.
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Following the pattern of the slope field through the point 𝐴, the solution does not coincide with the point 𝑆.
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So the point 𝐴 cannot belong to the same solution as 𝑆.
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Following the pattern through the point 𝐵, again this solution does not coincide with the point 𝑆.
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The same applies for point 𝐷 and also for point 𝐸, so that only point 𝐶 could belong to a solution containing the point 𝑆.
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We’ve seen how to draw a slope field and how to trace solutions through a point.
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Let’s look now at an example where we’re given the slope field and we want to find which differential equation slope field represents.
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Consider the given slope field graph.
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Which of the following differential equations is represented on the graph?
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A) 𝑦 prime is 𝑥 plus two over 𝑥 minus three.
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B) 𝑦 prime is two minus 𝑥 over 𝑥 minus three.
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C) 𝑦 prime is 𝑥 minus two over 𝑥 plus three.
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D) 𝑦 prime is 𝑥 plus two over three minus 𝑥.
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Or E) 𝑦 prime is equal to 𝑥 minus two over 𝑥 minus three.
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We can see that each of the options has a positive or negative two in the numerator and a positive or negative three in the denominator.
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And if we look at our graph, the behavior of the slope at 𝑥 is negative two and 𝑥 is positive three is quite distinctive.
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At 𝑥 is negative two, the slope field line segments are all horizontal.
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This means that the slope is equal to zero.
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That is, 𝑦 prime is equal to zero at 𝑥 is negative two.
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So let’s try 𝑥 is negative two in each of our possible equations and see if we get a match.
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At 𝑥 is negative two in our equation A, 𝑦 prime is equal to negative two plus two over negative two plus three.
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That’s equal to zero over negative five, which is equal to zero.
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So equation A at the point 𝑥 is negative two does actually match our slope field.
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If we look at equation B at 𝑥 is negative two, we have 𝑦 prime is equal to two minus negative two over two minus three.
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That’s equal to four over negative five, which is not equal to zero.
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So equation B does not match our slope field for 𝑥 is negative two.
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And we can eliminate equation B.
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Now let’s try equation C.
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For equation C, we have 𝑦 prime is equal to negative two minus two over negative two plus three.
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That’s equal to negative four over one, which is not equal to zero.
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So we can eliminate equation C.
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At 𝑥 is negative two in equation D, we have 𝑦 prime is negative two plus two over three minus negative two.
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That’s equal to zero over five, which is equal to zero.
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Equation D does match the slope field as 𝑥 is negative two.
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So equation D remains a contender.
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In equation E, we have 𝑦 prime is equal to negative two minus two over negative two minus three.
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That gives us negative four over negative five, which is four over five.
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This is not equal to zero.
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So it does not match the slope field.
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And we can eliminate equation E.
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We’re left now with equations A and D as possible options.
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Let’s now try 𝑥 is equal to positive three in each of these equations.
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At 𝑥 is equal to positive three, both equations have a denominator of zero.
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This means that both equations are undefined at 𝑥 is equal to three and have a vertical slope at that point.
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This corresponds to the slope on the slope field.
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So we’ll have to look elsewhere, for a possible solution.
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Let’s choose another value, say 𝑥 is equal to zero.
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At 𝑥 is equal to zero, each of the line segments in the slope field has a negative slope.
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So let’s see what the sign of the slope is for each of our two remaining differential equations at 𝑥 is equal to zero.
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In equation A, 𝑦 prime is equal to zero plus two over zero minus three, which is equal to two over negative three and is less than zero.
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This corresponds to the slope at 𝑥 is equal to zero in our slope field.
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So equation A is still a contender.
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In equation D, 𝑦 prime is equal to zero plus two over three minus zero.
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That’s equal to two over three, which is positive.
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And this does not correspond to the direction of the slope at 𝑥 is equal to zero in our slope field.
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So we can eliminate equation D.
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Only equation A remains.
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So the slope field graph represents the differential equation 𝑦 prime is 𝑥 plus two over 𝑥 minus three.
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In this example, we were given a slope field.
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And we had to fit the appropriate differential equation to the graph.
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In our next example, we start with a differential equation and need to find which of the given graphs matches the differential equation.
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Which of the following is the slope field of the differential equation 𝑦 prime is equal to two 𝑥 plus three 𝑦 minus five.
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The way we’re going to approach this is by a process of elimination.
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First, we’ll find a point on each graph, where 𝑦 prime, the slope, is equal to zero.
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That’s a flat line segment.
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If our differential equation does not equal to zero at such a point, then we can eliminate the associated graph.
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If we start with graph A, d𝑦 by d𝑥 is equal to zero at the point one, negative one.
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So the slope in graph A is equal to zero at the point one, negative one.
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Now, if we substitute 𝑥 is equal to one and 𝑦 is negative one in our differential equation, we have 𝑦 prime is equal to two times one plus three times negative one minus five.
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That’s equal to two minus three minus five, which is negative six.
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This is nonzero.
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So straight away, we can eliminate graph A.
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Now let’s look at graph B.
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Graph B has a slope of zero when 𝑥 is negative one and 𝑦 is negative one.
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If we try this in our equation, we have 𝑦 prime is equal to two times negative one plus three times negative one minus five.
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That’s equal to negative two minus three minus five, which is equal to negative 10.
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This is nonzero, so we can now eliminate graph B.
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Now let’s look at graph C.
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One of the points in graph C, where d𝑦 by d𝑥 is equal to zero, is the point three, two.
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With 𝑥 is three and 𝑦 is two in our equation, we have 𝑦 prime is two times three plus three times two minus five.
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That’s equal to seven, which is nonzero.
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And so we can eliminate equation C.
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Remember, we’re trying to find a graph with the point where the slope is equal to zero and where that point makes our differential equation also equal to zero.
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Let’s look at graph D.
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The slope is equal to zero in graph D at the point one, one.
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Substituting 𝑥 equal to one and 𝑦 equal to one into our differential equation, we have 𝑦 prime is equal to two times one plus three times one minus five.
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That’s equal to two plus three minus five, which is equal to zero.
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This matches with the slope on the graph at the point one, one.
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So graph D is a contender.
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Now, let’s look at graph E.
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We can see that, in graph E, at the point one, one, we have a slope of zero.
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We know already from graph D that the point one, one satisfies our differential equation.
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So let’s try another point on graph E which has a slope of zero.
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The slope is equal to zero in graph E at the point four, minus one.
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So let’s try this in our differential equation.
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We have 𝑦 prime is two times four plus three times negative one minus five.
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That’s eight minus three minus five, which is equal to zero.
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So far then, graph E also matches our differential equation.
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We now have two contenders, graph D and graph E.
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So let’s look again at graph D.
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In graph D, the slope is equal to zero at the point one, one.
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And this matches with our differential equation.
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The slope is also equal to zero at the point three, negative two.
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So let’s try this point in our differential equation.
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We have 𝑦 prime is two times three plus three times negative two minus five.
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That gives us six minus six minus five.
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And that’s equal to negative five, which is not zero.
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So we can now eliminate graph D.
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The only graph out of the five that could possibly match our differential equation now is graph E.
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Let’s just try another couple of points in graph E with slope of zero and see if they match our differential equation.
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The slope on the graph of zero at the point four, negative one and in fact 𝑦 prime in the differential equation is also equal to zero.
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The slope on the graph is equal to zero at the point negative two, three.
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And 𝑦 prime is also equal to zero at the point negative two, three.
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Out of our five graphs then, we can say that graph E is the only graph that matches the differential equation 𝑦 prime is two 𝑥 plus three 𝑦 minus five.
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Let’s just summarize now what we’ve seen.
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For a first-order differential equation of the form d𝑦 by d𝑥 is equal to 𝑓 of 𝑥, 𝑦 or d𝑦 by d𝑥 is equal to 𝑓 of 𝑥, or d𝑦 by d𝑥 is equal to 𝑓 of 𝑦, we can visualize a general solution without actually solving for the function 𝑦.
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We can do this by drawing a slope field.
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To do this, we draw our line segments, each with the slope d𝑦 by d𝑥 at various points 𝑥, 𝑦 in the 𝑥𝑦-plane.
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From this, we can get an idea of the pattern of the general solution of the differential equation.
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And with an initial condition, which is a starting point 𝑥 zero, 𝑦 zero, we can trace a particular solution through the slope field following the direction of the slope of the line segments.