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Explain why the functions with the given graphs can't be solutions of the differential equation$$\frac{d y}{d t}=e^{t}(y-1)^{2}$$

If $y=1, \frac{d y}{d t}=0 .$ However, the graph doesn't have a horizontal tangent line

Calculus 2 / BC

Chapter 7

Differential Equations

Section 1

Modeling with Differential Equations

Campbell University

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University of Michigan - Ann Arbor

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Lectures

01:11

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Explain why the graph of t…

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01:10

Give a graphical explanati…

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01:13

02:07

in this video, we're going to be looking at the graphs of solutions to differential equations. The differential equation we're looking at is why prime equals either the tee times the quantity y minus one squared. And you want to see if either of these graphs are potential solutions to this function uh to this differential equation. Sorry. So um by looking at this, let's first try and find the equilibrium solutions um to this differential equation. So any global solution occurs when y prime schools syrup. And so just my inspection, we can see that either the T can never equal zero. So the only way for this to occur would be when why minus one? I'm sorry. Why minus one equals zero. And that would be when y equals one. So a solution to this differential equation should have a slope of zero at y equals one. No matter what. Let's look at these two graphs. Um graph be first does not have a slope equal to zero at one, so no why prime equals zero at why? Of course one. So this cannot be a solution. Um graph A, however, does have a slope of zero at Y equals one, so, yes, why prime equals zero at my one. However, it also has this section right here, there's another equilibrium solution at a higher. Um Why value? And this cannot occur for this function? Because remember, the only way for why prime to equal zero is when why equals one? So why prime was zero? When why does not equal one? So this cannot be a solution either. So our conclusion is that neither neither graph can B a solution.

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