Question

Hi everyone! :o) Could somebody please tell me if I did anything wrong with my integration? I supplied a picture...thanks!

Answer 1

The reason I'm asking is because I think I have the wrong answer...i believe the answer should be y/x or 1/x(y) same thing

Answer 2

@ganeshie8

Answer 3

@zepdrix

Answer 4

this is actually the left hand side of a differential equation fyi...i wanted to work it out the hard way instead of using the shortcut of factoring out the d/dx method if that makes any sense

Answer 5

What's the other side of the equation look like? >.<

Answer 6

sinx dx
which turns into -cosx

Answer 7

if you want i could rewrite it like they did, that way you might better understand how and why I wanted to see if I could solve it without using the shortcut...but it will take me awhile to draw it out

Answer 8

whats the actual full problem ?

Answer 9

actually you can see the video here>>>
http://www.youtube.com/watch?v=p8yesVjNdzY
just skip through it

Answer 10

i was just trying to prove to myself that i could get the same answer without using the shortcut but so far, I'm not having any luck :o(

Answer 11

f(x) dy = g(y) dx

you cannot just integrate both sides and conclude

f(x)*y = g(y)*x + C

Answer 12

above is WRONG.

Answer 13

can you help me figure out how to do it without factoring out the d/dx step?

Answer 14

You don't like the Integrating Factor method? :o

Answer 15

no it's not that...i used the integrating factor to get to that point...i just don't want to factor out the d/dx..i wanna do it the long way for the learning experience but i need help

Answer 16

basically, i'm trying to make as many mistakes as possible so I can get my techniques running smoothly and prove to myself i am doing things the correct way...as well as prove to myself that i am good enough to mix up stuff if i choose to as well

Answer 17

so far, i kinda stink! :o)~

Answer 18

you're in right path trying out things :)

so your starting DE is
\[\large xy'-y = x^2\sin x\]
?

Answer 19

yeah...essentially

Answer 20

my integrating factor is 1/x

Answer 21

i multiplied both sides by 1/x

Answer 22

and then multiplied both sides by dx

Answer 23

that is where my attempt began at the left side of the equation if you click up above at my picture

Answer 24

The integrating factor can be obtained once you have the equation in y'+py=q form.

Answer 25

Where p and q are just a function of x

Answer 26

Then you proceed to write
vy'+vpy=vq

where derivative of v is suppose to by vp

so you can use the product rule
which states (vy)'=vy'+v'y
So we have
(vy)'=vq

Answer 27

where v is suppose to be your integrating factor of course

Answer 28

Do you know how to proceed from (vy)'=vq ?

Answer 29

i am writing it all out right now...i will post a pic in a few minutes...sorry i'm sooooo slow :o(

Answer 30

The problem seems to be that y is dependent with respect to x, so you cannot take functions of x out of the integral like constants. Instead, I think you have to treat this like such:
\( \displaystyle \int \frac1x \dfrac{dy}{dx} \ dx \)
and apply integration by parts. (dv = dy/dx dx, so that v = y)
The end result seems to be that the new integral piece cancels with the other integral. I suspect that is due to how integration by parts is related to the product rule as well.

Answer 31

i.e.
\( \dfrac{1}{x} y' - \dfrac{1}{x^2} y = \sin x \)
\( \displaystyle \dfrac1x \dfrac{dy}{dx} \ dx - \dfrac{1}{x^2} y \ dx = \sin x \ dx\)
\( \displaystyle \int \dfrac1x \dfrac{dy}{dx} \ dx - \int \dfrac{1}{x^2} y \ dx = \int \sin x \ dx\)
Letting u = 1/x, du = -1/x^2 dx; dv = dy/dx dx, v = y;
\( \displaystyle \dfrac1x y - \int y \left(-\dfrac{1}{x^2}\right) \ dx - \int \dfrac{1}{x^2} y \ dx = - \cos x \)
If you pull out the negative sign in the first integration term, you just get the opposite of the second, so they cancel out.

Answer 32

quick question...on the integral 1/x^2 y dx...you don't treat the "y" as a constant and pull it out front?

Answer 33

Nope, and for the same reasons. y is a function of x, so it does count for integrating with respect to x.
The end result really just cancels the two integrals of 1/x^2 y dx, so you don't actually have to do work there after integration by parts.

Answer 34

so would you say my silly little brain is somehow mixing up the rules of partial derivatives with what looks like but isn't a "partial integral"? lol

Answer 35

Like, if you actually plugged in y for these integrals, you would have some function of x (y = x sin x + C x, I think it was), and you wouldn't pull that out of the integral, right?

Answer 36

I think so. But considering how often we use x and y for things I can't blame you. :P
It is mainly the context here. If it were actually a multivariate problem (using x and y for independent variables), taking the variables out makes sense. But here y is just a function of x, so we can't have that liberty.

Answer 37

ok I think I understand...so im not totally crazy...there are actually things called partial integrals then?

Answer 38

Not quite; the most basic way I can think of this is, you have single variable cases where y depends on x. So we can define dy in terms of a function of x, dx.
In the case of multiple variables where x and y are independent of each other, dx and dy also are not defined in terms of one another. So y has no effect on the integral with respect to x, and vice versa.

Answer 39

that makes sense...so would that kind of scenario maybe be found if you were trying to like track the gravitational forces of like 3 bodies in space relative to each other and were actually crazy enough to try to integrate them analytically instead of numerically with a computer? :o/

Answer 40

I'd say it is possible, although I've never done something quite like that thus far! lol

Answer 41

yikes! 8-O

Answer 42

thank you sooo much for your help!!! what it be okay if I became your fan and maybe bother you in the future? :o)

Answer 43

I don't mind, although I am not online as much as I used to be. I was just wandering a bit on here again before college started for me. :)

Answer 44

i have one last question...no math though...

Answer 45

when I took all my calculus, it seemed 90% of calculus is algebra...am i correct in assuming that 90% of differential equations is integration techniques?

Answer 46

There is a good amount of integration involved (at least from what I remember while learning DE).

Early on you usually always have an integral involved with the solution. Later there are some exceptions (like, the exponential function makes for some interesting works in constant coefficient situations for any order of DE).
The more notable cases later on are usually things like the transforms (Laplace Transform and Fourier Transform have integral definitions) used for solving boundary value problems. That's all I can think of right now!

Answer 47

Sounds like im in for alot of work!

Answer 48

Yes, DE can be a good amount of work! I really enjoyed the subject though; I learned it on my own and not through a class, reading through a book called Advanced Engineering Math (I don't remember the author / edition off the top of my head).

Answer 49

so the integral 1/x dy...the answer is y/x + C...why do we treat 1/x as a constant in this one?

Answer 50

The way I keep it is: \( \displaystyle \int \frac1x \dfrac{dy}{dx} \ dx \)
When applying integration by parts, you do get y/x from u*v but then you subtract off the integral of v du, which was \( \displaystyle \int -\dfrac1{x^2} y dx \). This just happens to cancel with the other integral involved (well, its entirely because of the integrating factor being there), leaving y/x + C.

Answer 51

so what you're saying is that i just found an example that just happens to look like you treat 1/x as a constant but it isn't always like that...essentially that example fools you into thinking you treat the integral sort of like a partial derivitive, but its mathematical quicksand?

Answer 52

I think it will always look like that, because of how the situations are always set up. I'll write this in general form, using a very lax notation (to save time)
\( y' + Py = Q \)
\( My' + MP y = Q \) By definition of integrating factors, M' = MP
\( My' + M' y = Q \)
\( \displaystyle \int My' \ dx + \int M' y \ dx = \int Q \ dx \)
Apply integration by parts to the first integral.
u = M, du = M' dx; dv = y' dx, v = y.
\( \displaystyle My - \int M' y \ dx + \int M' y \ dx = \int Q \ dx \)
\( \displaystyle My = \int Q \ dx \)

But integration by parts is really just another form of the product rule for derivatives, which is where the 'shortcut' comes from.
My connection to OS is being weird today!

Answer 53

* Oops, I forgot to multiply the right side by M. So that right side is always "MQ". The left side is still the same though.

Answer 54

how do you remember all these little details so well?

Answer 55

they make you write down so many rules in all these math classes, there's no way i could sit and rewrite all that

Answer 56

I've had a lot of practice with the rules, and I guess a lot of patience too!

For me the best way to memorize things has been to just repeatedly write down the rule whenever I used it until it stuck. It's just been mechanical from there.

Answer 57

good advice! okay well thanks for all the info and help...i better get back to work now...thanks again! :o)

Answer 58

Yup! Good luck!
I've also considered using flash cards for help in memorizing a lot of the rules, but I've never actually done it for myself. So maybe that could help out as well.

Answer 59

good idea!...thanks Access

Answer 60

No problem!