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Homework Help: Ordinary Differential equation

  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Note: Here your solution is implicitly defined, i.e. you can not rearrange the solution to get an explicit expression for y. Therefore you need to enter your solution as an equation. You should enter your solution in the form f(x,y) = constant where you have determined both f(x,y) and the constant (for example: sin(y)/x=ln(2) ). You must include an equals sign. Do not use decimal numbers.



    2. Relevant equations
    (5+7y^2)dy/dx +(4x+2)y = 0, y(1)=1


    3. The attempt at a solution
    had a try at using integrating factor but having the equation = 0 doesnt really help
    then tried breaking them up into seperable equations
    ((5+y^2)/y)dy = (-4x-2)dx
    integrated and got f(x,y)=5lny+y^2-2x^2-2x+9/2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 1, 2010 #2
    (5+y^2)/y = (5/y) + y
    when you integrate (5/y) + y , you should get 5Iny + (1/2)y^2
    and when you shift your intergrated x components to the left side, shouldnt it be positive?
    thats what i think..haha
     
  4. Apr 1, 2010 #3
    hmm yea tried that aswell
    also tried integrating:
    (5+y^2)dy = (-4xy-y)dx
    got 5y +(y^3)/3 = -4x^2y-2xy+c
    with y(1)=1
    got c = 34/3
    but still wrong >.>
     
  5. Apr 1, 2010 #4
    i dont think you can do it that way. You cant simply multiply the y over and integrate wrt x and treat y as constant.

    opps! and sorry i read ur qns wrongly.
    (5+7y^2)dy/dx +(4x+2)y = 0

    this is 3rd degree DE? think i haven learn this yet..haha..sorry!
    =P
     
  6. Apr 1, 2010 #5
    its a first degree DE since it has dy/dx in it, not d^3y/dx^3
    at least to my knowledge
     
  7. Apr 1, 2010 #6

    Mark44

    Staff: Mentor

    That's a good start, but you made a mistake on the left side.
    You should have (5/y + 7y)dy = (-4x - 2)dx

    This is not quite right. You should have something like 5 ln(y) + <other x and y terms> = <constant>.
     
  8. Apr 1, 2010 #7

    Mark44

    Staff: Mentor

    There is no "In" function. That's Ln, which is sort of an abbreviation of natural Logarithm.
     
  9. Apr 1, 2010 #8
    oh wow i cant believe i mistook the question, i didnt see 7y^2 while trying to work it out maybe that will help
     
  10. Apr 1, 2010 #9
    this is the answer i got:
    5logy + 7/2y^2 + 2x^2 + 2x + 15/2
    but it still did not work
     
  11. Apr 2, 2010 #10

    Mark44

    Staff: Mentor

    That's because you lost your equation. Remember what I said...
    Starting from here
    (5/y + 7y)dy = (-4x - 2)dx
    Integrate both sides of this equation to get
    5lny + (7/2)y^2 = -2x^2 -2x + C

    Now use the initial condition y(1) = 1 to find C, and you're done, so stop at that point.

    If you want to check your answer, differentiate implicitly and you should get back to the original differential equation.
     
  12. Apr 2, 2010 #11
    yes using the initial conditions
    5ln1 + 7/2(1)^2 + 2(1)^1 +2(1) = C
    which is
    5(0) + 7/2 + 2 + 2 = C
    C = 15/2 doesnt it?

    which is the same equation i got and it still doesnt work
     
  13. Apr 2, 2010 #12

    Mark44

    Staff: Mentor

    Yes, I got C = 15/2. Why do you think it doesn't work?
     
  14. Apr 2, 2010 #13
    have tried entering it on the computer and it still is not right

    sorry it was right, instead of entering "ln(y)", i entered as "lny" damn syntax
     
    Last edited: Apr 2, 2010
  15. Apr 2, 2010 #14

    Mark44

    Staff: Mentor

    Keep that in mind. We take a lot of shortcuts in writing math expressions, that a computer program is not likely to understand, especially with trig functions and log functions. Instead of cosx or sinx, it's probably going to be looking for cos(x) and sin(x).
     
  16. Apr 2, 2010 #15
    yep thanks for you help :)
     
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