Question

Exact ODEs
I need some help pls

Answer 1

I called the first term M and the second N.
Then I calculated:
\[M_y = \frac{15 e^{-x} \cos (5y)}{x}-\frac{2x}{y^2}\\N_x = 15e^{-x}\cos(5y)-\frac{2}{y^2}\]

I think\[R = \frac{1}{N}(M_y - N_x)\], but somehow it doesn't feel like that would work?

Answer 2

@phi @Loser66 care to help?

Answer 3

have to go to school now friend, sorry, wait for phi.

Answer 4

@phi

Answer 5

off hand, I don't remember the details...

Answer 6

Aww:( Thanks, anyway

Answer 7

:)

Answer 8

up, over, down, and up again ...

Answer 9

not sure how the thrm goes, but if i were to take a stab at it ...

exact odes have the property that Fxy = Fyx

\[F_{xy}=-\frac{2x}{y^2}e^{u}-\frac {15}xe^{-x+u}cos(5y)\]
\[F_{yx}=-\frac{2}{y^2}e^{u}-\frac{2x}{y^2}u'e^{u}-{15}(-1+u')e^{-x+u}cos(5y)\]
and compare parts ....

Answer 10

I attached a pic of the theorems.
They just say how to get the integrating factor R.


Referring to the original question, I need to determine a, b and c

Answer 11

Oh not the integrating factor (that is F). The theorem says how to get R, which is used for calculating F

Answer 12

its referencing a (16) which is not in the picture, which might help me see the difference between F(x) and F(y)

Answer 13

\[R=\frac{1}{Q}(P_{xy}-Q_{yx})\]

Answer 14

In your previous answer, why did you use second orders? It is P_y - Q_x ?

Answer 15

just a misreading on my part :) i was thinking P and Q were already in Px and Qy forms for some reason

Answer 16

I got this question from a previous year's lecturer.
It's from a matlab assignment where you run a script and it asks you some questions.

I just took a look at the code:
A=input('a = ? \n');
B=input('b = ? \n');
C=input('c = ? \n');

b1='1';b2='1';b3='x';

istop = 0; marksFact = 3;

while A~=b1 || B~=b2 || C~=b3 ;

istop = istop + 1; marksFact = marksFact - 1;

if istop == 3;
disp('You failed this part 3 times and are awarded no marks here!')
marksFact = 0;
break;
else;
disp('At least one of a, b or c is wrong; try again')
A=input('a = ? \n');
B=input('b = ? \n');
C=input('c = ? \n');
end

It looks as if a =1, b = 1, c = x.

Maybe we can work backwards, from that?

Answer 17

does this look like the correct setup so far? latex and mathing dont mix to well :)
\[R=\frac{1}{-\frac{2x}{y^2}-15e^{-x}cos(5y)}\left(-\frac{2x}{y^2}+\frac {15}xe^{-x}cos(5y)+\frac{2}{y^2}-15e^{-x}cos(5y)\right)\]

Answer 18

I think that's what I got as well, I'll check quickly

Answer 19

Yip I got that as well.
When I got that, I came to openstudy to ask the question, because I don't really see how that expression can help us get the a/b/c as asked?

Answer 20

http://www.wolframalpha.com/input/?i=%281%2F%28%28-2x%29%2Fy%5E2+-15+e%5E-x+cos%285y%29%29%29%28%28%2815+e%5E-x+cos%285y%29%29%2F%28x%29-%282x%29%2Fy%5E2%29-%2815+e%5E-x+cos%285y%29-2%2F%28y%5E2%29%29%29

Answer 21

the rest is just algebra, pain staking algebra at that

\[\frac{-\frac{2x}{y^2}+\frac {15}xe^{-x}cos(5y)}{{-\frac{2x}{y^2}-15e^{-x}cos(5y)}}+\frac{\frac{2}{y^2}-15e^{-x}cos(5y)}{{-\frac{2x}{y^2}-15e^{-x}cos(5y)}}\]

\[\frac{-2x+\frac {15}xy^2e^{-x}cos(5y)}{{-2x-15y^2e^{-x}cos(5y)}}+\frac{2-15y^2e^{-x}cos(5y)}{{-2x-15y^2e^{-x}cos(5y)}}\]

\[\frac{-2x+\frac {15}xy^2e^{-x}cos(5y)}{{-2x-15y^2e^{-x}cos(5y)}}+\frac{2-15y^2e^{-x}cos(5y)}{{-2x-15y^2e^{-x}cos(5y)}}\]

Answer 22

and they want it in some sort of: a - (b/c) form?

Answer 23

It looks so, that's what confused me.... I have no idea what to do

Answer 24

im not sure of the steps they took, but it looks to me like the first "results" on the wolf is:\[1-\frac{1}{x}\]

Answer 25

if we subtract that from our getup we get 0

http://www.wolframalpha.com/input/?i=%281%2F%28%28-2x%29%2Fy%5E2+-15+e%5E-x+cos%285y%29%29%29%28%28%2815+e%5E-x+cos%285y%29%29%2F%28x%29-%282x%29%2Fy%5E2%29-%2815+e%5E-x+cos%285y%29-2%2F%28y%5E2%29%29%29-%281-1%2Fx%29

Answer 26

1-1/x agrees with what I go from the scripts code...

Answer 27

then id prolly go with that :)

R(x) = 1 - 1/x

Answer 28

So then a = 1, b =1, c = x hahah, whatever they did

Answer 29

That is quite a stupid answer to ask students to master exact, don't you think?

Answer 30

im sure there was something else along the lines that would have made it simpler .... but im not aware of all the thrms and such ....

Answer 31

or, they like to torture you with algebra ;)

Answer 32

hahahah

Answer 33

Thanks for the help!

Answer 34

youre welcome, and good luck