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Math Help - first order DE show that

  1. #1
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    first order DE show that

    show that the general solution of

    dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)

    and express a, b and w in terms of p,q and an arbritrary constant C

    I believe I have to integrate as differentiating y will not produce a constant so

    let px-qy=v

    dy/dx=(p-dv/dx)/q

    dv/dx=p-qe^v=p(1-(q/p)e^v)

    seperate variables

    int(p/p-qe^v)dv=int(p)dx

    v-ln(p-qe^v)+c=px

    qy= c-ln(p-qexp(px-qy))

    How to get in required form?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poirot View Post
    show that the general solution of

    dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)

    and express a, b and w in terms of p,q and an arbritrary constant C

    I believe I have to integrate as differentiating y will not produce a constant so

    let px-qy=v

    dy/dx=(p-dv/dx)/q

    dv/dx=p-qe^v=p(1-(q/p)e^v)

    seperate variables

    int(p/p-qe^v)dv=int(p)dx

    v-ln(p-qe^v)+c=px

    qy= c-ln(p-qexp(px-qy))

    How to get in required form?
    I haven't looked at your substitution method in any detail, but it might work. However I think you are overthinking this. The equation separates nicely in its original form.

    -Dan
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  3. #3
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    Which equation are you referring to ? I was thing about the law ln(ab)=ln(a) +ln(b) but unfortunatly it is a sum, not a product that is being logged.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poirot View Post
    dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)
    \displaystyle \frac{dy}{dx} = e^{px - qy}

    \frac{dy}{dx} = e^{px}e^{-qy}

    e^{qy}dy = e^{px}dx

    -Dan
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  5. #5
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    Yes I realised last night.
    After integration, I get 1/q(e^qy)+c=1/pe^px,

    y=1/q(ln(q/p(e^px)-qc))

    w= -qc (which I know is another constant but I thought it best to be consistent)
    a=1/q, b=q/p

    Thanks for your help
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