Question

Hi everyone! I would like to discuss a few questions regarding theory of solution curve generation. Just a couple of questions to clear thing up for me. Thanks! :o)

Answer 1

@hartnn @Kainui

Answer 2

hi i will help you

Answer 3

yes

Answer 4

When learning how to draw solution curves in class, my professor showed taught us to set our DE equal to "C"..."C" being the slope...

Answer 5

Now this confused some people because people are getting the "C" from constant of integration mixed up with the "C" from simply the slope you want your drawn around the isocline you generate...do I have the concept correct that the "C" you set the DE equal to has NOTHING to do with the "C" from the constant of integration you get after solving an indefinite integral?

Answer 6

this constant can be anything and it is different however the reason for constant in a DE is also Integration while solving. @SinginDaCalc2Blues

Answer 7

sorry ribhu, can't really follow what you wrote here

Answer 8

let me be specific...

Answer 9

the constant C in both the DE and Integration comes. they need not to be same.

Answer 10

and in DE, we solve it using integration so there has to come a constant of integration.

Answer 11

in your comment above "this constant can be anything and it is different however the reason for constant in a DE is also Integration while solving"...when you say "this constant of integration" what specifically are you referring to? The "C" you choose to set your DE equal to, or the "C" you get after solving an indefinite integral?

Answer 12

@SinginDaCalc2Blues Why don't we work through a simple example and see for ourselves? \[\Large \frac{dy}{dx} = x\] Let's say that's our differential equation, try the method your teacher describes and also solve it and see if the constants are related some how.

Answer 13

in a line y=mx+c, slope is m and intercept is c which is an intercept.

Answer 14

@SinginDaCalc2Blues this can be explained it simply doesn't matter that is just a way to represent constant. in a DE constant can be calculated using Initial Conditions. and in indefinite integral it is not needed to be calculated. there is simply no need for this comparision.

Answer 15

okay Kainui...so you want me to solve it analytically for "y"?

Answer 16

\[\int\limits\limits_{}x^2{}dx = x^3/3+C\]

Answer 17

y=(1/2)x+C

Answer 18

where as you are given that dy/dx = x^2
which will give you y = x^3 / 3 +C both are C but the value of C will depend on the variety of initial conditions present.

Answer 19

Well I gave \[\Large \frac{dy}{dx}=x \] so the answer should be \[\Large y=\frac{1}{2}x^2+C\] I think @SinginDaCalc2Blues you just forgot the exponent since you put the 1/2. Alright cool so now do it the way your teacher did it, but use something else like K to show it's a different constant (unless we can somehow show they're the same) You with me so far? =P

Answer 20

YEAH...SORRY i FORGOT THE ^2

Answer 21

and apparently I am SCREAMING for some reason lol

Answer 22

Haha no big deal XD

Answer 23

ok...lemme draw the curves...brb

Answer 24

i suggest you invest in some chain mail and pillows...the beatings are going to hurt...and sorry, I don't give out my email...if you have a question, please post it but not inside my question please

Answer 25

well if I set dy/dx equal to K...then x=K...but for some reason I am confused, maybe because there is no "y" in the DE...not sure why I am confused, it should be easier with less variables...ugh

Answer 26

Well we could change the function maybe to y'=y*x or something? Or maybe can you show me a specific example he worked in class?

Answer 27

yes...I've got one...y'=y-2x
I got that one from here>>> https://www.youtube.com/watch?v=hsTfC1Wmxqc
She is working the problem in a different way than I was taught...it looks like she is doing it like a computer would, not a human...click the link and fast forward to like 7:30 just to look real quick and you will see what I am talking about

Answer 28

she isn't even drawing isoclines, just lineal elements and then drawing in her solution curve at the end...seems like a harder way to do it

Answer 29

Yeah that's the way I learned how to do it originally too, but I don't think it matters too much. So the answer to this problem is: \[\Large y=2x+2+Ce^x\] and you can at least check that it's true because it will satisfy the differential equation. So maybe now you can compare this to the other constant you're using to see if they're the same or not?

Answer 30

okay well just let me reiterate what you want me to do to see if I understand...

Answer 31

Sure, yeah I know I sort of magically pulled this solution out of thin air, but the point is to compare the constants we're using to see if they're the same or not, not really how we got them if that makes sense.

Answer 32

If I do it the way I learned in class, I will simply set y-2x=C, but you want me to do y-2x=K instead right?

Answer 33

Yes exactly. =)

Answer 34

okay well what's confusing me a little bit is if I do it my way, I will simply choose different values for K, like somewhere between -4 and +4 for example...so basically I am choosing the slope of the lineal elements I want all around each isocline I create(I'm just talking my way through this if you couldn't tell) :o~

then...

Answer 35

then I choose x and y values for each block of "K's" I choose so I can complete the isoclines...then I draw the isoclines...then last, I simply sketch in to the best of my ability the solution curves...that sums up my method...so now focusing on her method...

Answer 36

Sure that's fine, sounds good. I think you understand what "isocline" means right? "iso" means the same and "cline" means slope, so really you're just looking at the lines on the graph where they have the exact same slope. This is different than the constant of integration, since that will shift the graph up and down. But what we're doing here is showing that they are different if that makes sense. It's just hard to explain.

Answer 37

For the "computer method without isoclines", I just simply choose x and y values and they generate a y' value, and I can sketch a lineal element at that specific point...after I do that for like a gazillion points, I draw my solution curve with no need for isoclines...

Answer 38

so now that I have talked my way through both procedures, I haven't been able to figure out how to do what you asked me to do...ugh sorry lol

Answer 39

oh yes, just read what you wrote and that makes sense!

Answer 40

actually I thought of a neat way to explain it to myself earlier...at least I think I did...

Answer 41

That's fine, so for instance I said the answer to the differential equation\[\Large y'=y-2x\] was this: \[\Large y= 2x+2+Ce^x\] So how do I prove it? Just take the derivative and plug it in: \[\Large y' = 2+Ce^x\] So plug this in \[\Large y'=y-2x \\ \Large 2+Ce^x = 2x+2+Ce^x-2x \\ \Large \text{ true!}\] But remember what you did was set \[\Large y-2x=K \\ \Large \text{ same as} \\ \Large y' = K\] and we can see that
\[\Large K \ne 2+Ce^x\] since those are both the derivative, so they're different! =P

Answer 42

when I took geology, we had lots of maps with "contour lines" on them that basically connected points that had the same elevation...now those weren't "isocline" but they do demonstrate the concept really well of curves drawn on paper that connect similar things...so since "iso" means the same and "cline" means like a slope/gradient/inclination/declination/etc etc, an "isocline" simply means "connecting all the points on the graph that have the same slope! :o) right?

Answer 43

Yeah exactly! Those contour lines on the map were all the points that had the same height, not the same slope, so very similar, good analogy!

Answer 44

yes...so the "C" from the integration has NOTHING to do with the "C" from the isocline procedure! lol

Answer 45

Yeah "cline" as in I like to relax and put my feet up while sitting in a recliner. =P

Answer 46

Or climbing a steep incline to the summit, etc etc... haha.

Answer 47

ok whew!...they totally SHOULD NOT USE "C"!!! They should use something else like K because it's too confusing!

Answer 48

so my intuition was correct then at the beginning then...so that's good!

Answer 49

I had one more quick question about a concept but I have to remember what it was so I need a sec to gather my thoughts...lol

Answer 50

Haha well perhaps, but really C is just a letter like x or y or z or F or Q or anything else. Really there is no meaning in "C" it's just a letter. It's kind of like arguing we should use C in the word "Cooking" because "Cat" already has a C in it and we don't want to confuse people into thinking we cook cats! All I'm saying is try to remove that sort of bias if you plan on taking higher math courses because it's not really useful to assume the meaning of a letter, you should try to put the meaning into the letter based on the context. =P

Answer 51

I agree but when you complete an integral one minute, and a new concept of solution curves shoves a "C" down your throat, you get confused thinking you use that "C" for something else or it comes from an integral when it doesn't

Answer 52

and that gets confusing to the novice mathemetician

Answer 53

True, I definitely agree with you, but your math teacher probably forgets sometimes, but at least now you know haha.

Answer 54

I guess my last question was a simple one...should one method be used over the other? computer vs human method? the computer method's graph is much cleaner without the isoclines all over but when you do the human method with the isoclines, it's really quick to sketch in the lineal elements with all the same slope really quick...what do you think?

Answer 55

Hmm, I think it depends on the problem. Personally I wouldn't have done either since I knew how to solve it with an integrating factor. Whichever you think is easier or whichever one your teacher wants you to do on the test/homework I suppose.

Answer 56

well I think this section is teaching us how to draw slope fields so we aren't supposed to solve analytically right now

Answer 57

I learned about the human/computer method from Walter Lewin on the MIT videos...have you seen those?

Answer 58

Yeah I've seen several of his, he's pretty good, I like a lot of the stuff off MIT OCW, pretty awesome. The internet is amazing haha.

Answer 59

if you look at exactly 3:00 on this vid>>>
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-1-the-geometrical-view-of-y-f-x-y/

he says one you can't solve...do you think you can help me solve the one that you can solve? I think it might just be a simple linear DE ...its y'=y-x^2

Answer 60

Oh I might be able to solve that y'=y-x^2

Answer 61

I can't seem to get it into Linear form (dy/dx)+P(x)y=f(x)

Answer 62

I only know Bernoulli and Linear and homgenous and seperable...I thought it might be linear but I am not having much luck :O(

Answer 63

Just subtract y from both sides, you'll have f(x)=-x^2 and p(x)=-1

Answer 64

Integrating factor s a lot of fun imo and it gives you another way to look at the product rule which is cool.

Answer 65

This is really an interesting topic. Although I thought that the C (constant) in integration is discussed thoroughly even before covering applications of integration.

Answer 66

OMG!!!! why didn't I see that!? geez

Answer 67

you are probably right nincompoop, but we forget things :o)...omg how we forget things! :o)

Answer 68

ok I am going to finish solving brb

Answer 69

integrating factor e^(-x)...still going...

Answer 70

Yeah good so far. =)

Answer 71

brb gonna grab a snack

Answer 72

quick q...after you mult both sides by your int factor, do you do your reverse product rule at that point? I keep thinking I am suppose to divide but I may be confusing linear with bernoulli?

Answer 73

wait...reverse product rule, then divide, then integrate I believe...lemme try that

Answer 74

y=-x^2

Answer 75

Not quite, although you are on the right path. Can you type out your steps?

Answer 76

omg...wait...I totally forgot the C...after all that talk of C, how can I forget the C! UGH...brb

Answer 77

actually I'm confused...I never even integrated because after multiplying both sides by e^-x, then I did reverse product rule which gave me ye^-x = -x^2e^-x...
then I just multiplied both sides by e^x without even integrating because I saw by just multiplying both sides by e^x, I would get y=...is this wrong methodology?

Answer 78

I HAVE TO INTEGRATE no matter what don't I?

Answer 79

EVEN IF IT LOOKS LIKE IT'S FINISHED, IT REALLY ISN'T IS IT?

Answer 80

sorry, yelling again

Answer 81

if I integrate rather than cancel the e^-x, will I have to do Int. by Parts once on the left and twice on the right?

Answer 82

lol I was talking with my parents in the other room about what we want to make for breakfast, so I'm sort of in and out at the moment but would you like me to walk you through an example, like how I did the one before this, y'=y-2x since it's really similar?

Answer 83

I think you're almost right on the method but you've sort of skipped some steps. I'll make it clear what's going on every step of the way, since it actually all makes a lot of sense.

Answer 84

you don't need to walk me through it, just list the steps is fine...1)put in Standard form(linear) 2)find integrating factor, 3)multiply both sides by integrating factor 4)do reverse product rule on the left side, 5)then what? lol

Answer 85

Then after that you integrate both sides, then divide both sides by the integrating factor.

Answer 86

okay so 5) integrate both sides
so that means integration by part once on the left and twice on the right then?

Answer 87

\[\Large u y' +u'y =u f(x) \\ \Large (uy)' = u f(x) \\ \Large uy = \int\limits u f(x)dx \\ \Large y = \frac{1}{u}\int\limits u f(x)dx\]

Answer 88

when integrating the left side, since its with respect to x, the y is a constant so...
-ye^-x is the left side...I will put the C on the other side...brb

Answer 89

No, that's not true at all, y is a function of x! It is definitely dependent. What we are saying is: \[\Large (uy)' = \frac{d}{dx} (uy)\] so when we integrate both sides with respect to x we are really saying: \[\Large \int\limits (uy)' dx = uy + C\]

Answer 90

ok...my final answer is y=x^2-x+1-Ce^x

Answer 91

is that right?

Answer 92

Looks like you messed up a few constants, just on the -x+1 should be 2x+2, but everything else is perfect.
http://www.wolframalpha.com/input/?i=y%27%3Dy-x%5E2&t=crmtb01

Answer 93

oh yeah...i foprgot the 1/2...geez...one sec

Answer 94

y=x^2-2x+2-Ce^x

Answer 95

is the -Ce^x okay or does it have to be positive?

Answer 96

What is C?

Answer 97

a constant