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mathslover
 one year ago
Solve the given differential equation.
\[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y2)}\]
mathslover
 one year ago
Solve the given differential equation. \[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y2)}\]

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mathslover
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 @Australopithecus @robtobey

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3What methods have you learned? I don't think this is separable

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2I tried putting y = vx and x + y = v but none of them seem to work.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3First bring everything to one side and make it equal to zero

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2\(\cfrac{(x+2)(y2) dy}{(x+y)^2 dx}  1 = 0 \) What next?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3That is in incorrect form

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3You want the form: \[M(x,y)dx + N(x,y)dy = 0\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3You can use substitution or you can check to see if it is exact and use that method

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Substitution is as follows: You have two choices for substitution: u = x/y then, dx = udy + ydu Or u = y/x then, dy = xdu + udx

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3For setting it in the right form do you understand?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, I've studied the method of substitution. In fact, I did try substituting y = vx but that didn't seem to work.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3You wrote it in the wrong form so that could be your problem

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2I understand that point. But, am not sure where to put that \((x+y)^2\) ... \((y2)dy = \cfrac{dx}{x+2} \times (x+y)^2\)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3So, for example: \[\frac{dx}{dy} = \frac{(y + x)}{(x + 2y)^2}\] in standard form would be: \[dx(x+2y)^2  (y+x)dy = 0\] M(x,y) = (x + 2y)^2 N(x,y) = (y+x)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Sorry this would be standard form \[(x+2y)^2dx  (y+x)dy = 0\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Okay, so, I have : \((y2)(x+2) dy  (x+y)^2 dx = 0 \) This is the standard form, right?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Ok now you have to test to see if you can even use the separation method

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Or substitution sorry

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3This is clearly not seperable

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Now, I will simply use substitution : y = vx \(dy = vdx + xdv \) The equation becomes: \((vx  2)(x+2)(vdx + xdv) (x+vx)^2 dx = 0 \) \((vx2)(x+2)(vdx + xdv) = x^2 (v+1)^2 dx \) Should I proceed further or am I doing something wrong?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Expand (y2)(x+2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Also expand (x+y)^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3MY browser is lagging like crazy one second

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3there is a test you have to do to make sure you can even use substitution method

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Okay, it will give me : \[(y2)(x+2) dy  (x+y)^2 dx = 0 \\ xydy + 2(yx) dy  4dy  x^2 dx y^2 dx 2xy dx = 0 \\ xydy  2y dy  2xdy  x^2 dx  y^2 dx  2xydx = 0 \]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Test? Never heard of it. Can you please explain?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Now factor out dy and dx

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2\[(xy  2y  2x)dy  (x^2 + y^2 + 2xy)dx = 0 \\ (xy  2y  2x)dy = (x^2 + y^2 + 2xy)dx \] Seems like I'm reverting everything back to the original equation :/ :(

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3You can only do substitution method if \[M(\lambda x, \lambda y) = \lambda^pM(x,y)\ for\ all\ \lambda,x,y\] and \[N(\lambda x, \lambda y) = \lambda^qN(x,y)\ for\ all\ \lambda,x,y\] to use separation p must equal q

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3First to check if M(x,y) or N(x,y) is homogenous here are some examples:

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3The degree of homogeneity must be equal for both terms N(x,y) and M(x,y) to use the substitution method

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3its a pretty simple check you can do in your mind almost

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2I seem to get your point! Will give it a try next morning. Gotta sleep now! :) Thanks a lot for your help! I'll let you know if I get stuck anywhere.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Wait lets at least check to see

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3because you might have to use another method

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Is M(x,y) homogenous?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3and to what degree

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Sorry I know you are probably tired just I want to save you time

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3If you are lost I dont mind giving you another example

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2It doesn't seem to be homogeneous :( because of that constant there

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3So you have to use a different method

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Oh! :( Any ideas for which method to use?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3You can check to see if it is exact

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Do you want to sleep we can work on this later if you want?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3its actually a pretty long method

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Oh! Well, I will love to give it a try now. I can control my sleep for studies!

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Alright, first step you need to test to see if the DE is exact or not

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3A DE is said to be exact when, \[M_y(x,y) = N_x(x,y)\] You know how to take partial derivatives right?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Also I would put it in another form but I am not sure how to use write daran in latex

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3So first step take both partial derivatives and check to see if they are equal

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3if they are not this question is even longer ha

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2\(\delta\) You mean this?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3that is lower case delta

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3dw:1432921439849:dw

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3anyways we are getting off topic

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3Have you taken the partial derivatives? You should always take both of them because you will always need them to solve this kind of problem

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2I will do that part. What I'm thinking right now is that this question appeared in my book before the topic of "Exact DE" ... and it wil surely involve the method of solving H.D.E

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3I Dont know the HDE method

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2I meant, solving Homogeneous Differential Equations

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3oh im out of my mind lol

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3anyways this is the I would use to solve this problem

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3it should work

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3I mean unless you make a mistake with algebra substitution method wont work with this

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Well, I guess, I might have made a mistake then. I will see it tomorrow morning and will let you know about the progress, if you don't mind. As I'm getting a bit out of mind too :(

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3That's fine just message me or tag me or something I can keep going over this with you, This is essentially a really brief note on how to solve this problem IF it is exact

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.3If you need help later have a nice sleep

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2:) Thanks a lot for that, @Australopithecus . And same to you (whenever you schedule to sleep :D )

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Sleep is important ! Have one now, please! :D

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Well, I got it! We had to substitute x + 2 = u and y 2 = v and then work out with y = kx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The ODE as given isn't homogeneous. For \(\dfrac{dy}{dx}=f(x,y)\) to be considered homogeneous, \(f\) has to satisfy \(f(tx,ty)=t^kf(x,y)\) for some real \(k\). \[\begin{align*}f(x,y)&= \frac{(x+y)^2}{(x+2)(y2)}\\\\ f(tx,ty)&= \frac{(tx+ty)^2}{(tx+2)(ty2)}\\ &= t^2\frac{(x+y)^2}{(tx+2)(ty2)}\\ &\neq t^kf(x,y) \end{align*}\] which explains why you were running into trouble with the substitution \(y=vx\). As for your substitution, I was actually in the process of suggesting that! Nice work.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Yeah! :) I don't know how did it click just when I woke up ! :D Indeed, I solved it in 5 minutes :P As compared to hour of work last night. And thanks for that explanation, @SithsAndGiggles ! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sometimes the hardest thing is remembering that \(22=0\) :P

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Haha! Agreed. This is what "Maths" is...! Happens in programming too, when the error is ";" .. and we keep finding it throughout the code :D It's fun...
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