Problem

For what values of $ r $ does the function $ y = …

02:02

Question

Answered step-by-step

Problem 75 Hard Difficulty

Show that the function $ y = e^{2x} (A \cos 3x + B \sin 3x) $ satisfies the differential equation $ y" - 4y' + 13y = 0. $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Anjali Kurse

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Video Transcript

Okay. So we have this function and we want to show that it satisfies this differential equation. So how do we prove that? What we have to do is take a bunch of derivatives. So we have this function? Mhm. Because I had three X plus B signed three X. Mhm. Okay. And what we have to do here is take the derivatives because this Y double prime minus four Y. Prime plus 13 Y. It has Y double prime and why prime? Which are derivatives of? Why? So one thing we can do to satisfy to prove that this function satisfies this differential equation is plug in the derivatives into this equation and see if it comes out zero. So let's do that. We're gonna need to find why prime and why the whole prime for that. So my first step is going to be to distribute this either of the two X. Just to get rid of these parentheses. So we have a nice clean function to look at. So I'm going to bring this constant A upfront. So it looks a little bit standardized. Mhm. Okay, so this turns into A E. To the two X. Co sign three X plus B. E. Two the two X. Sign three X. So now let's take our derivatives. We're going to need to use the product rule here. So what is product rule? Well, when we have two functions with X in them we need to use product rule which means a prime times B plus B prime times A. So let's do that. Why equals if we have if this part is A And this part is B. Let's use this product rule A. Prime. The derivative of E. To the two X. Is going to be to A E. To the two X. Okay. Times be we'll just actually I don't need this parentheses be we're going to keep be exactly the same. Plus be prime times A. So what is be prime be prime? The derivative of cosine three X. Is going to be negative three sign three X. Okay, I'll use my parentheses here and then we'll keep A. The same so we'll keep E to the two X. Cool. Now we got to do the same thing for this second one. So this is number one. This is number two. Okay so let's do that. This will be our A. And this function signed three X. Will be R. B. So when we do that, the derivative of A. Is to be E. To the two X. Times side three X. Because we have a prime to be eat the two X. And we keep be exactly the same. So it's just signed three X. Plus. Now we're going to take the derivative of B. The derivative of B. Is three. Co sign three X. And we're going to multiply this by A. So just be E. To the two X. So we had to use the product rule here because we have two functions with exit them into the two X. And coastline three X. In both of these terms. So that's why we had to use product role and that is how we do that. This is why prime. So now let's just simplify this. We have to a Co Sign three X. Can we reduce any terms here? Let's see. We have this to be signed three X. But the other side term has an A. Not to be so we can't combine those but maybe we can put them in a slightly better order. So let's do that To a key to the two x Co signed three X. So if we just want to put the co sides together then we'll put this term next. We'll get three B. E. Two the two X. Co sign three X plus to await should do are a term next. If we want to keep with this A. V. Order and this isn't if another if you have another variation of this with the terms in a different order, it doesn't matter too much. As long as you have the pluses and minuses and the terms correct. This is just so it's a little standardized to be each of the two X. Sign three X. Okay, there we go. No, let's do our next step. Which is going to be why double prime And this is going to have four product rules. So much fun. Okay, quick reminder. Product rule is a prime times B plus B. Prime times A. Okay, So we're gonna have four product rules to do here. Okay, the derivative of to A. E. To the two X. Is going to be four A. E. To the two X. Then we're going to keep our be termed the same. Okay then we're going to add our next term. Yeah. Which is going to be, yeah, the prime times A. So what's the derivative of coastline? Three X. It's -3 sign three x. And then we keep the A term the same. So we'll get to A. E. To the two X. Here. Okay, so that takes care of this first one. Let's do the next one. Okay. The derivative of A. This is going to be A. Is six B. Eat to the two X. This is going to be multiplied by the B term kept exactly the same. Okay, and then let's add, let's add um be prime times A. Be prime is negative three sign three X. And A. Is three b. e. two. The 2 x. Uh huh. Okay. Just To make this 100% clear, let's switch to that. Okay, next we're making progress minus this is going to be this whole term here. So I don't have to deal with this minus minus a. Prime Is going to be the derivative of 382. The three x. Each of the two X. Sorry, which is going to be six. A key to the two X. We keep our beach be term exactly the same. Okay, And then we have our be prime Which is going to be three sign three X then. Sorry this is getting a little squished here. See if I can Scroll over just a little bit. three signed 3 x times. Sorry that should be three co signed three acts. If we're taking the derivative of the B term here, the derivative of this term three co sign three x. times three A. E. To the two X. Okay, we'll clean that up in a bit. But let's do this fourth term first. Mhm. Okay. So we have that fourth term, the derivative of the first part is going to be 4B. E. To the two X. We keep the second part, the B. Part the same then mhm. We take the derivative of the second part part which is going to be three cosign 3 X. And we multiply it by the first part which is the same. Okay jesus is some ugly math here. Lots of algebra. Okay, so now if we just completely right this out, let's see if any any terms will cancel six A. E. To the two X. Sign three X. All I'm doing here is clarifying this whole thing. So our first term is for a each of the two exco same three X. Right here then we have negative six A. To the two X. Side three X. Then we have plus let's see six B. E. To the two X. Co sign three X minus this part minus nine. The E. To the two X. Sign three x minus. So much algebra minus six A. E. To the two X signed three X minus. Because we want to distribute this negative here, distribute this negative minus nine A. E. To the two X. Co sign three X. All I did here is multiply these three these threes times each other and bring then bring the nine to the front And then plus four b. E. To the two X. Sign three X. Finally plus six B. Each of the two X. Co sign the reacts. Okay so now let's find the terms that might be in common. This one has a B and a co sign. So we can add those. This one let's see this one has A and a co sign. This one has an A. And a casa no one has an and assign silly this one has an A. And a co sign. Okay, this one has an A and a sign and a sign that leaves these to be and sign be inside and all I'm doing here is trying to group these like terms so that we can simplify this equation a little bit further. So finally we get Y double prime is equal to take the greens together. We're gonna get -5 a. E. To the two X. Co sign 33 X. And the reason I can do this is because the E. To the two X. And the co sign three X. Are the same in these. So we can really just um subtract the coefficients and see what happens. And then we'll do our reds. So this is our green should have used that same color, sorry. And then our reds are negative six A. Eight to the two X. And three X. And negative six A. Each of the two X. And three X. O. Negative 12 either of the two X. Sign three X. And then this is our reds. And then our blues are going to be six B. E. To the six B. E. Two. The two X. Co signed three X plus six B. E. Two the two X. Co signed three X. 0. 12 B. E. To the two X. Co sign three X. And then finally the ones underlined in black negative nine B. Each of the two X. And three X plus four B. You did the two X. And three X. O plus no minus negative nine plus negative 90 plus four B. Is negative five B. E. To the two X. Sign three X. Okay, there we go. Now our job is to put this Y. Prime with our Y. Prime which is here with our Y. And plug them into this defeat Keogh basically and see if this is true. So we have our Y double prime here. Let's just erase this blocks around them honestly. Yeah. Okay. So if we have this right? Mhm. Plus Plus four times. Why prime? Let's just plug in. Ry prime here. Are Y. Prime is to A. E. To the two X. Yeah. Uh Co sign three x plus three B. Each of the two X. Co sign three X. Three B. To the two X. Co sign three X Plus -3 A. Eat to the two x. sign three x. Plus Plus to be each of the two x. sign three X. Sorry I forgot to be there. Come on mm slight technical difficulties here. Okay Plus to be each of the two X. Sign three acts Okay so we have that just make sure that looks right here. Okay And then we have this last one minus 13. I'm sorry it should be a -4 here minus four plus 13. I'm going to change this color to black because that's whatever they were doing plus 13 times. What's the original? E. A. E. To the two X. Co sign three X. Plus you B. Each of the two x. Sign three X. Okay so now what we have to do is evaluate this. So let's see what we can do here. So let's write this out negative five sorry negative five E. To the two X. Co sign three X minus 12. Each of the two X. Sign three X plus 12 E. To the two X. Co sign three x minus five. B. E. Two. The two X. Sign three X. Okay and then we'll add this. Why? Prime term? Which is what this is minus Here. I'll just move this minus down here -8 a. You did the two x. Co sign three X. Remember we're distributing this negative four minus 12 B. Either the two x. Co signed three x. Plus 12 A. Eat of the two X. Sign three X -8. Be either of the two x. Sign three X. Okay and then our last one is going to be plus 13 82. The two X. Co sign three X. Plus 13. Yes. 13 B. E. Two the two X. Sign three X. Okay this should be equal to zero. So let's underline our life terms here. We've got E. To the coastline. Three X. 82. The two X. And three X. We have this one and this one so look at that negative five A. Negative eight am positive 13 cancel out to zero. Next we have negative 12 E. To the two X. Sign three X. Sorry I forget that A. Or B. They're my bad. Um Oh 12 B. If we look at this That should be 12 a. Okay yep. 12. Okay sorry about that. Sorry about that. And how I found that was the -6 -6. I must have just forgotten to write that a. In there. So let's make space for that. You got 12 A. E. to the two x. Okay So that's our green. Where else do we have that positive? 12 E. To the sign three X. Cool. So that adds up to zero. Next 12 E. To the two X. Co sign three X. Well we have negative 12 B. E. Two. The two X. Coastline three X. Those cancel it to zero. Okay. Last but not least. We have our negative five B. E. Two. The two X. And three X. And we have our negative eight be either of the two X. And three X. That makes negative 13. And then when we had this one in 13 B. To the two X. 93 X. When we add all those up we get zero. So I know this was a really messy problem with lots of algebra but this is how you do this. To prove the differential equation is equal to zero. You have to calculate why prime and Y double prime. Something that could result in a lot of mistakes as you almost saw with my A. There. But this is how you solve this kind of problem

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

For what values of $ r $ does the function $ y = …

This textbook answer is only visible when subscribed! Please subscribe to view the answer

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Caleb Elmore

Samuel Hannah

Michael Jacobsen

Joseph Lentino

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Caleb Elmore

Samuel Hannah

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Caleb Elmore

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Caleb Elmore

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals