mathslover
  • mathslover
Solve the given differential equation. \[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

mathslover
  • mathslover
Solve the given differential equation. \[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathslover
  • mathslover
Australopithecus
  • Australopithecus
What methods have you learned? I don't think this is separable
mathslover
  • mathslover
I tried putting y = vx and x + y = v but none of them seem to work.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Australopithecus
  • Australopithecus
First bring everything to one side and make it equal to zero
mathslover
  • mathslover
\(\cfrac{(x+2)(y-2) dy}{(x+y)^2 dx} - 1 = 0 \) What next?
Australopithecus
  • Australopithecus
That is in incorrect form
Australopithecus
  • Australopithecus
You want the form: \[M(x,y)dx + N(x,y)dy = 0\]
Australopithecus
  • Australopithecus
You can use substitution or you can check to see if it is exact and use that method
Australopithecus
  • Australopithecus
Substitution 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
  • Australopithecus
For setting it in the right form do you understand?
mathslover
  • mathslover
Yeah, I've studied the method of substitution. In fact, I did try substituting y = vx but that didn't seem to work.
Australopithecus
  • Australopithecus
You wrote it in the wrong form so that could be your problem
mathslover
  • mathslover
I understand that point. But, am not sure where to put that \((x+y)^2\) ... \((y-2)dy = \cfrac{dx}{x+2} \times (x+y)^2\)
Australopithecus
  • Australopithecus
So, 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
  • Australopithecus
Sorry this would be standard form \[(x+2y)^2dx - (y+x)dy = 0\]
mathslover
  • mathslover
Okay, so, I have : \((y-2)(x+2) dy - (x+y)^2 dx = 0 \) This is the standard form, right?
Australopithecus
  • Australopithecus
Yup :D
Australopithecus
  • Australopithecus
Ok now you have to test to see if you can even use the separation method
Australopithecus
  • Australopithecus
Or substitution sorry
Australopithecus
  • Australopithecus
This is clearly not seperable
mathslover
  • mathslover
Now, I will simply use substitution : y = vx \(dy = vdx + xdv \) The equation becomes: \((vx - 2)(x+2)(vdx + xdv) -(x+vx)^2 dx = 0 \) \((vx-2)(x+2)(vdx + xdv) = x^2 (v+1)^2 dx \) Should I proceed further or am I doing something wrong?
Australopithecus
  • Australopithecus
Expand (y-2)(x+2)
Australopithecus
  • Australopithecus
Also expand (x+y)^2
Australopithecus
  • Australopithecus
MY browser is lagging like crazy one second
Australopithecus
  • Australopithecus
there is a test you have to do to make sure you can even use substitution method
mathslover
  • mathslover
Okay, it will give me : \[(y-2)(x+2) dy - (x+y)^2 dx = 0 \\ xydy + 2(y-x) dy - 4dy - x^2 dx -y^2 dx -2xy dx = 0 \\ xydy - 2y dy - 2xdy - x^2 dx - y^2 dx - 2xydx = 0 \]
mathslover
  • mathslover
Test? Never heard of it. Can you please explain?
Australopithecus
  • Australopithecus
Now factor out dy and dx
mathslover
  • mathslover
\[(xy - 2y - 2x)dy - (x^2 + y^2 + 2xy)dx = 0 \\ (xy - 2y - 2x)dy = (x^2 + y^2 + 2xy)dx \] Seems like I'm reverting everything back to the original equation :/ :(
Australopithecus
  • Australopithecus
You 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
  • Australopithecus
First to check if M(x,y) or N(x,y) is homogenous here are some examples:
Australopithecus
  • Australopithecus
The degree of homogeneity must be equal for both terms N(x,y) and M(x,y) to use the substitution method
Australopithecus
  • Australopithecus
hence p = q
Australopithecus
  • Australopithecus
its a pretty simple check you can do in your mind almost
mathslover
  • mathslover
I 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
  • Australopithecus
Wait lets at least check to see
Australopithecus
  • Australopithecus
because you might have to use another method
mathslover
  • mathslover
Yeah sure!
Australopithecus
  • Australopithecus
Is M(x,y) homogenous?
Australopithecus
  • Australopithecus
and to what degree
Australopithecus
  • Australopithecus
Sorry I know you are probably tired just I want to save you time
Australopithecus
  • Australopithecus
If you are lost I dont mind giving you another example
mathslover
  • mathslover
It doesn't seem to be homogeneous :( because of that constant there
Australopithecus
  • Australopithecus
So you have to use a different method
mathslover
  • mathslover
Oh! :( Any ideas for which method to use?
Australopithecus
  • Australopithecus
Yup
Australopithecus
  • Australopithecus
You can check to see if it is exact
Australopithecus
  • Australopithecus
Do you want to sleep we can work on this later if you want?
Australopithecus
  • Australopithecus
its actually a pretty long method
Australopithecus
  • Australopithecus
kind of
mathslover
  • mathslover
Oh! Well, I will love to give it a try now. I can control my sleep for studies!
Australopithecus
  • Australopithecus
Alright, first step you need to test to see if the DE is exact or not
Australopithecus
  • Australopithecus
A DE is said to be exact when, \[M_y(x,y) = N_x(x,y)\] You know how to take partial derivatives right?
mathslover
  • mathslover
Yeah!
Australopithecus
  • Australopithecus
Also I would put it in another form but I am not sure how to use write daran in latex
Australopithecus
  • Australopithecus
So first step take both partial derivatives and check to see if they are equal
Australopithecus
  • Australopithecus
if they are not this question is even longer ha
mathslover
  • mathslover
\(\delta\) You mean this?
Australopithecus
  • Australopithecus
that is lower case delta
mathslover
  • mathslover
\(\Delta\) .. ?
Australopithecus
  • Australopithecus
|dw:1432921439849:dw|
Australopithecus
  • Australopithecus
anyways we are getting off topic
Australopithecus
  • Australopithecus
Have 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
  • mathslover
I 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
  • Australopithecus
I Dont know the HDE method
mathslover
  • mathslover
I meant, solving Homogeneous Differential Equations
Australopithecus
  • Australopithecus
oh im out of my mind lol
Australopithecus
  • Australopithecus
anyways this is the I would use to solve this problem
Australopithecus
  • Australopithecus
it should work
Australopithecus
  • Australopithecus
I mean unless you make a mistake with algebra substitution method wont work with this
mathslover
  • mathslover
Well, 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
  • Australopithecus
That'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
  • Australopithecus
If you need help later have a nice sleep
mathslover
  • mathslover
:) Thanks a lot for that, @Australopithecus . And same to you (whenever you schedule to sleep :D )
Australopithecus
  • Australopithecus
It is never
mathslover
  • mathslover
Sleep is important ! Have one now, please! :D
mathslover
  • mathslover
Well, I got it! We had to substitute x + 2 = u and y -2 = v and then work out with y = kx
anonymous
  • anonymous
The 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 \(k\). \[\begin{align*}f(x,y)&= \frac{(x+y)^2}{(x+2)(y-2)}\\\\ f(tx,ty)&= \frac{(tx+ty)^2}{(tx+2)(ty-2)}\\ &= t^2\frac{(x+y)^2}{(tx+2)(ty-2)}\\ &\neq t^kf(x,y) \end{align*}\] 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
  • mathslover
Yeah! :) 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
  • anonymous
Sometimes the hardest thing is remembering that \(2-2=0\) :P
mathslover
  • mathslover
Haha! 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...

Looking for something else?

Not the answer you are looking for? Search for more explanations.