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How to solve these Differential Equations?

  1. Apr 7, 2005 #1
    Hi,

    Q 1. ( e ^ y/x - y/x e ^ y/x + 1 / 1 + x^2 ) dx + e ^ y/x dy = 0

    Ans. I know its Homogeneous sub, y = ux,
    then, dy / dx = u dx + x du

    I did this, and got to the point,

    e^u dx + 1 / ( 1 + x^2) dx + x. e^u du = 0

    How can we separate this now?

    Q 2. y dx + ( 2x - y e^y ) dy = 0

    I think we can use exactness, here

    M = y

    My = 1

    N = 2x - y e^y

    Nx = 2

    Not exact,

    so,

    Integrating Factor would be : e ^ Integral My - Nx / N

    Is this right. Integrating factor is getting to complicated to multiply the eqn, with. Any better way of doing this.


    Q 3. ( 2x + tany) dx + ( x - x^2 tany ) dy = 0

    I think here too, exactness, may be used, but any better way, if possible.
     
  2. jcsd
  3. Apr 7, 2005 #2

    dextercioby

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    Science Advisor
    Homework Helper

    The second one is really simple...

    [tex]\frac{dx}{dy}=\frac{ye^{y}-2x}{y} [/tex]

    Nonhomogenous linear ODE.The homogenous one is separable.

    Daniel.
     
  4. Apr 7, 2005 #3
    Alright, let's write these out:

    1. [tex]\left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0[/tex]

    2. [tex]y \ dx + (2x - ye^y) \ dy = 0[/tex]

    3. [tex](2x + \tan y) \ dx + (x - x^2\tan y) \ dy = 0[/tex]

    #1 is exact (check for yourself!). It isn't homogeneous.

    An integrating factor that will work for #2 is just [itex]y[/itex] (you can determine it by inspection...).

    Integrating factor isn't that bad for #3 (although you will definitely know which function to integrate after you find it...).
     
  5. Apr 11, 2005 #4
    So, far had now luck with part 1,

    If it is exact,

    I tried to find partial derivatives with respect to y and x for both terms, and then applied

    My - Nx / N

    and the Integrating factor gets really messy....

    I got 2 ,,, piece of cake.

    3. if M = 2x + tany
    then

    My = sec^2y

    N = x - x^2 tany

    Nx = 1 -2xtany

    Then,

    Integrating factor = Nx - My / M

    1-2xtany - sec^2y / 2x tany

    Trig, identity,

    sec^2y - tan^2y = 1
    sec^2y = 1 + tan^2y

    Plugging in this into: 1-2xtany - sec^2y / 2x tany , for sec^2y

    we get,

    1-2xtany -1 -tan^2y / 2x tany

    After simplifying the above, I got:

    Integrating factor as - tany

    When I try to multiply both sides by -tany, the expression doesn't get any simpler, but infact becomes harder to evaluate,

    Any ideas, on how this could be made simpler.

    Thanks
     
  6. Apr 12, 2005 #5
    1. How the heck to find the partial for no. 1 with respect to x and y.....

    It is a pain. I would appreciate if somebody could help me with number 1 and 3


    Thanks
     
  7. Apr 12, 2005 #6
    [tex]\left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0[/tex]

    I can't help you with solving the equation but the partials:

    [tex] M_y = \frac{e^{\frac{y}{x}}}{x} - \frac{e^{\frac{y}{x}}}{x}+\frac{ye^{\frac{y}{x}}}{x} = \frac{ye^{\frac{y}{x}}}{x}[/tex]

    [tex] N_x = \frac{-ye^{y/x}}{x^2}[/tex]
     
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