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(a) For what values of $r$ does the function $y=e^{r x}$ satisfythe differential equation $2 y^{\prime \prime}+y^{\prime}-y=0 ?$(b) If $r_{1}$ and $r_{2}$ are the values of $r$ that you found in part (a)show that every member of the family of functions$y=a e^{r_{1} x}+b e^{r_{2} x}$ is also a solution.

(a) $\frac{1}{2},-1$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 1

Modeling with Differential Equations

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Boston College

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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in this video, we're going to be looking at verifying solutions to a differential equation. So the first thing we are asked to do is um find all values of our um has seen this equation here. This is what defines our Y equals E. To the X. Such that why is a solution to the differential equation seen here? So in order to do this, what we're going to do is just plug this solution into the differential equation and see what values are, will make it a true state. So we know why and so we can plug Y directly in but we need to find why prime and why double prime. Let's go ahead and do that. So let's start with Y prime. Oh pumps why prime is going to equal um R. E to the R. X. Then if we take the derivative of that again to find Y double prime, we will get y double prime equals R squared eat to the R X. So if we take those and are expression for why and plug those in, What we will end up with is mhm. Two times R squared E to the Rx uh Plus yeah R E to the R X minus E. To the Rx. That all equals zero. So let's go ahead and simplify this. We get two R squared E. To the X plus R. E. To the Rx minus. Eat there, we'll see her. So there's a need of the Rx and all three of these terms. We can go ahead and cancel that out and now we're left with two R squared plus are minus one equals zero. So now in order to solve this equation for our, we can do that by factoring so if we go ahead and factor this out, we chat are plus one times two ar minus one equals zero. So we know we can now set these factors equal to zero. It's all for our and get our solutions to this equation so are Plus one equals 0. Um so if we saw that we get our equals negative one and let's get the other factor equal to 02 ar minus one equals zero. So then we get our equals one half, that is our other solution. So those are our two values for our um So for the next part of this question, we are asked to show that oh functions of the form Y equals A E D R one X plus B B to the R two X. Our solutions to this differential equation and um A and B here are constants and are one in our to refer to the two solutions that we found for our in the previous part of the question. Um so we can go ahead and call are 1 -1 and R two is one half. So um using that information we get why schools hey to the negative X. Was B. E. The X. Over to that. Those are the um That's the family of functions that we want to test whether or not it is a solution. So let's go ahead and plug that into our differential equation. Again we need to find um why prime and why double? So let's do that. Why prime equals negative A. Each of the negative X plus um Be over to each of the X over two. Then let's take the derivative again. White double prime. We get A. E. To the negative X plus be over four. Eat the X over two. And now we can plug into our differential equation. So two times Y double prime, it's going to get us to A. E. To the negative X plus B. Over to each of the X over two and then plus Y. Prime minus A. E. To the negative X plus be over to E. To the X over two and then minus Y equals zero. And I'm just going to continue this on the next line, so minus A. E. To the negative X minus B. E. To the X over two. And this has to equal zero. So let's see if this is a true statement. So if we go ahead and collect our A terms, we have to A. E. To the negative X minus A. Either the negative X minus a. Negative X. And those all cancel each other out that is zero. And then if we look at RB terms, we'll be over to eat the X. Over to a positive, you're to eat the eggs, X. Over to another positive. Be over to eat at the X. Or two and then a negative B, times E to the X over two. And those also all cancel each other out. So we end up with the true statement zero equals zero. So this family, oops. Yeah. Of functions is a solution.

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