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Solve the differential equation dy/dx=xy+y^2

  1. Aug 21, 2005 #1
    How to solve the differential equation
    dy/dx=xy+y^2

    In my case this leads to an integral which is unsolvable! :cry:
     
  2. jcsd
  3. Aug 21, 2005 #2

    arildno

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    Most ODE's are analytically unsolvable.
    It might be there exists a trick to manage this one, but off-hand, I can't help you.
     
  4. Aug 21, 2005 #3

    Zurtex

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    The answer can not be expressed in terms of elementary functions. Although with the use of the quite commonly used Erfi function there is a fairly simple representation of the answer, why don't you show us what you have done so far.
     
  5. Aug 21, 2005 #4
    I have done it this way:
    i have put -1/y=t =>1/y^2dy=dt

    dt/dx + xt =1
    Integrating factor comes out to be e^(x^2/2)
    So finally i have to integrate e^(x^2/2) to get the solution.!
    I can do the integral but what perplexes is me that it's solution is not in form of elementary functions!
     
  6. Aug 21, 2005 #5

    GCT

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    It's a Bernoulli equation, you can solve it by Leibniz substitution

    [tex]v=y^{(1-n)}[/tex]
     
  7. Aug 21, 2005 #6
    That's exactly what i have done,but later it is unsolvable,or rather solution is not clear to me.
     
  8. Aug 21, 2005 #7

    saltydog

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    Alright, this is how I see it starting from:

    [tex]d\left[e^{x^2/2}z\right]=-e^{x^2/2}[/tex]

    Integrating:

    [tex]\int_{x_0,z_0}^{x,z} d\left[e^{x^2/2}z\right]=-\int_{x_0}^x e^{x^2/2}dx[/tex]

    Yielding:

    [tex]z=Ke^{-x^2/2}-e^{-x^2/2}\int_{x_0}^x e^{t^2/2}dt[/tex]

    Noting that:


    [tex]\int_{x_0}^x f(t)dt=K+\int_0^x f(t)dt[/tex]

    We can write the above expression as:

    [tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\int_0^x e^{t^2/2}dt[/tex]

    Letting [itex]u=t/\sqrt{2}[/itex] we obtain:

    [tex]z(y)=Ce^{-x^2/2}-\sqrt{2}e^{-x^2/2}\int_0^{x/\sqrt{2}} e^{u^2}du\quad\tag{1}[/tex]

    Now, it just so happens that:

    [tex]Erfi[x]=\frac{Erf[ix]}{i}[/tex]

    and:

    [tex]Erf[ix]=\frac{2i}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

    so the i's cancel and we're left with:

    [tex]Erfi[x]=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

    or:

    [tex]\int_0^{x} e^{t^2}dt=\frac{\sqrt{\pi}}{2}Erfi[x][/tex]

    Substituting that into (1) we get what Mathematica reports:

    [tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\sqrt{\frac{\pi}{2}}
    Erfi\left[\frac{x}{\sqrt{2}}\right][/tex]

    But z is 1/y so then 1 over all that stuf is the answer (if I didn't make any mistakes). How about a plot?

    Also, I'll quote someone in here:

    "Equal rights for special functions"

    That means, treat Erfi[x] in the same way as Sin[x]. You wouldn't mind if the answer was:

    [tex]z(x)=Ce^{-x^2/2}Sin[x][/tex]

    would you? Same dif.
     
  9. Aug 21, 2005 #8

    saltydog

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    Oh yea, back-substitute everything into the ODE in Mathematica to make sure the answer is right. We can do that . . . like Gauss and the rest of them wouldn't have used Mathematica if they had it back then. Be for real. :smile:
     
  10. Aug 22, 2005 #9
    Thanks Salty :smile:
    Everything is fine and well for me in yours solution!
    Could you please throw some light on this New Emergent function!
    I came across this first time!
     
  11. Aug 22, 2005 #10

    saltydog

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    Well Erfi[x] is called the imaginary Error function and Erf[x] is the error function. Check out Mathworld for details. But just any valid expression can hold for a new function. There's tons of them. For example, if I have some irreducible integral of the form:

    [tex]2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

    I can call it Sal[x] such that:

    [tex]Sal[x]=2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

    and further:

    [tex]\frac{dSal[x]}{dx}=\frac{1}{\sqrt{ln(x)+c}}[/tex]


    Same dif as Sin[x] as far as I'm concerned.
     
    Last edited: Aug 22, 2005
  12. Aug 22, 2005 #11
    Thanks Salty! :smile:
     
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