A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

solve the following equation if it is exact: (dy/dx)=(2xy-3x^2-2)/(6y^2-x^2+3).... in my previous question, I wrote x^3 but it should have been x^2.

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what does it mean by solve? find the function that it was derived from?

  2. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it might have something to do with it being an homogeneous function..... but I am just learning about them :) if its homogeneous, you can substitute y=vx into the equation and separate the variables.... when you get to the end of it, replace v by v=y/x. or something along those lines

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Manipulate the equation so it is in Mdx+Ndy=0 form. To check if it's exact, partial M with respect to y has to equal partial N with respect to x. If they are equal, then you can apply the exact d.e method. Let F be another function where the solution to the d.e is F=c. Since the equation is exact, the following statements are true and will be applied. Partial F with respect to y equals M and partial F with respect to x equals N. The way I remember which N and M goes with which variable is to look at the dx and dy term. Since the first term is Mdx, it's partial F with respect to x since it corresponds to the dx. So now, since you have partial F with respect to x is equal to M. We can sub in our value of M since we know what it is. Our end goal is to find the function F, so we need to integrate with respect to x. Instead of just C, your constant will be a function of y, H(y) because you are integrating with respect to x. Now, to completely solve for the function F, you need H(y). So using your equation, partial F with respect to y equals N. Obtain partial F with respect to y by differentiating what you just got earlier with respect to y, keeping in mind your constant H(y) becomes H'(y). Now, equate this back to N. Rearrange using algebra to solve for H'(y). Integrate to get H(y). Sub H(y) back into your function F. Equate F=c and you are done.

  4. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah.... just like that :)

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when you say partial M with respect to x, do you mean take the partial derivative of the m function with respect to x while holding the y's constant?

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so if M(x,y)dx= 2xy-3x^2-2, dM/dx will be 2y-3?

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2y-3x*

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah that's right. An easier notation that is usually used is just Mx=2y-3x with the x being a subscript. Because dM/dx is different than the partial operator but I'm sure you know that.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i guess where I'm getting confused is when I test to see if it is exact which it should be, I am getting Mx=Ny, in this case, 2y-3x=12y which isn't a true statement

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's My=Mx, not other way around.

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My=Nx*

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so I got two equations, df/dx=m=2xy-3x^2-2 and df/dy=N=-6y+x^2-3. I don't really follow what you were saying from here. do I integrate both of these, set them equal to eachother and solve for a y or an x?

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, you only need to integrate one. It doesn't matter which one you choose, but I always choose to integrate M. Our goal is to find F remember, so since df/dx=M you have to integrate with respect to x and then there's the whole thing about the constant y ou need to be careful of which I wrote above. Then you get the function F after integrating. You want to differentiate with respect to x this newly obtained F to get df/dx. You are doing this because you need to solve for the constant. Equate that to N, and solve for the constant you got earlier.

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    after integrating m, I got f(x,y)= ((2xy-2)*x)-x^3+c. if I differentiate this, won't I just get the same function that I integrated?

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No because you are differentiating with respect to y now. Sorry I said x in the above post. And like I said, your constant is not just C. It is a function of y, H(y) because you integrated with respect to x, therefore any function of y also acts as a constant.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.