Question

How do you make this DE separable?

Answer 1

D E
Hope it helps

Answer 2

\[x*\frac{ dv }{ dx }= \frac{-w \sqrt{x^2+v^2}+v^2-v^2x}{vx}\]

Answer 3

Above is the equation that needs to be separated. How to please? I don't know what to do with he square root.

Answer 4

the*

Answer 5

couldn't you just square the whole equation? I'm not that good on these things tbh, I lost practice :P

Answer 6

Well I don't know if that's allowed too. I mean am I allowed to square dv/dx? Because it would be an easy question if I was. So if anyone can confirm I can or not, please share

Answer 7

you should be able too, but in any case, you'd have to square X too -> (x*dv/dx)^2 (?)

Answer 8

But once you do that do you integrate it twice then?

Answer 9

I guess... well, I'll refrain from giving my opinion from here on, since I could just end up confusing you more :$ sorry

Answer 10

And how do you integrate (dy)^2

Answer 11

Okay. Thank you anyways

Answer 12

Is this even possible?!

Answer 13

What I have right now is: \[vdv = [-w \sqrt{x^2+y^2} + v^2(1-x)]dx\]

Answer 14

So that's pretty much where I'm stuck.

Answer 15

So pissed off right now

Answer 16

i assume that "w" is a constant?

also it may be that this DE is simply not separable and another method needs to be used

Answer 17

Yes w is constant

Answer 18

Well the question says it is... ERRRRR

Answer 19

Want to kill someone right now lol

Answer 20

oh ok, i will keep looking at it ... sometimes it takes some fancy substitutions to make it work

Answer 21

So basically I can't square root both sides right?

Answer 22

I mean square both sides

Answer 23

squaring both sides will just complicate things

no you dont want the "dv/dx" squared

Answer 24

Yep I thought so I wouldn't know what to do when that happens anyways

Answer 25

So okay the original equation is this: http://s3.amazonaws.com/answer-board-image/cramster-equation-20089201527116335752123113180668984.gif

Answer 26

But I have made the v = y/x substitution

Answer 27

I'm not sure what to do with the square root. Its supposed to be separable once you substitute v = y/x

Answer 28

why is there a "v' in original DE?

are we solving for y(x)?

Answer 29

Yeah we're solving the equation by making a substitution y = v/x

Answer 30

The thing is when I do the substitution I can't even go to integral part because I cannot separate all of the variables

Answer 31

So basically this is what I'm doing: \[v + \frac{ dv }{ dx } =\frac{ -w \sqrt{x^2+v^2} + v^2 }{ vx }\]

Answer 32

And to explain more about your question about the original equation... "v" is constant and so is w

Answer 33

Here's the whole question: http://www.chegg.com/homework-help/questions-and-answers/differnetial-equations-question-group-projects-chapter-3-d-aircraft-guidance-crosswind-air-q3808571?cp=CHEGGFREESHIP

Answer 34

ok theres a problem, if "v" is a constant then dv/dx = 0
you cant make the substitution y = v/x because y(x) is the solution you are solving for.

Answer 35

I'm so confused because it specifically says make y=v/x substitution to solve the equation. This equation is correct: http://s3.amazonaws.com/answer-board-image/cramster-equation-20089201527116335752123113180668984.gif but how do you even substitute y=v/x in there and solve it?

Answer 36

Unless there;'s another substitution method instead of using y = v/x?

Answer 37

I'm so screwed in so many levels...

Answer 38

where does it say to make the substitution, it just seems odd to do that

i can easily show that "y=v/x" is not a solution to the original DE

Answer 39

Then how do you solve it then?

Answer 40

not sure yet...

Answer 41

Actually this link doesn't say that its separable but it does on my homework. http://www.chegg.com/homework-help/questions-and-answers/differnetial-equations-question-group-projects-chapter-3-d-aircraft-guidance-crosswind-air-q3808571?cp=CHEGGFREESHIP

Answer 42

hmm i think this maybe is what they were referring to .... the "v" is different than the constant v in this problem lol

http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

Answer 43

Assume
that the wind speed and the speed of the aircraft through the air (airspeed)
are constant. (This is dierent from ground speed).

Answer 44

So basically v = aircraft speed, w= wind speed

Answer 45

this is a homogeneous de right? http://s3.amazonaws.com/answer-board-image/cramster-equation-20089201527116335752123113180668984.gif

Answer 46

because it says that I should have obtained homogeneous de and then I can make the v=y/x substitution

Answer 47

right but use a different letter like "u" so to not confuse with velocity constant

Answer 48

Okay but I'm still stuck with the square root thing

Answer 49

\[\sqrt{x^2+v^2}\]

Answer 50

it goes away once you make the substitution

u = y/x ----> y = ux

Answer 51

you did something wrong

\[\sqrt{x^2 + y^2}\]
y = ux
\[\sqrt{x^2 + u^2 x^2}\]
\[x \sqrt{1 + u^2}\]

Answer 52

after substituting "y=ux" you should get this:
\[u + x \frac{du}{dx} = u - \frac{w \sqrt{1+u^2}}{v}\]

Answer 53

is it separable now with that equation?

Answer 54

yep

Answer 55

Because this is what I'm getting: \[\frac{ dux^2 }={ \frac{ -wx \sqrt{1+u^2-vu(1-x)} }{ vx } }\]

Answer 56

Whoops forgot to put the dx on the RHS

Answer 57

Errr that formula messed it up hold on not that one