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Cannot prove the DE is exact?

  1. May 24, 2012 #1
    Cannot prove the DE is exact??

    Hey guys Im looking at a non-linear first order DE. the problem is : (y^2)/2+2*y*exp^(x) +(y+exp^(x))dy/dx=exp^(-x) y(0)=1. I put everything on the same side: ((y^2)/2+2*y*exp^(x)-exp^(-x))dx+(y+exp^(x))dy=0. This equation is not exact so I use (My-Nx)/N and got a function of x alone that equaled 1. Put it into exp^(∫1dx)=exp^(x); took this and multiplied it my N and still M=(y^2)/2+2*y*exp^(x)-exp^(-x) and the "new" N=exp^(2*x)+y*exp^(x): and still My≠Nx. It is still not exact and i have no idea where to go from here???? help
     
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  3. May 24, 2012 #2

    vela

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    Re: Cannot prove the DE is exact??

    Please use the template when posting here, and try to make your post easier to read. I did a little reformatting and clean-up for you.

    You don't multiply N and not M by the integrating factor. You have to multiply both because you're multiplying the equation by the integrating factor.
    $$e^x\left(\frac{y^2}{2} + 2ye^x-e^{-x}\right)\,dx + e^x(y+e^x)\,dy = 0$$ If you check the new M and N in that equation, you'll find it's now exact.
     
  4. May 25, 2012 #3
    Re: Cannot prove the DE is exact??

    Yeap... that was a rookie mistake on my part. I do have another question: I have solved the problem down to the point where [itex]∂F/∂x[/itex]=(2ye2x+y2ex-2x)/2+A(y) and [itex]∂F/∂y[/itex]=ex(y+ex)→e2x+A'(y) am I correct in answering A'(y)=yex because it is the only variable of y in [itex]∂F/∂y[/itex] which i would then intergrate to find A(y)=(y2ex)/2 + C and put this back in the equation for [itex]∂F/∂x[/itex] and solve to the constant C?
     
  5. May 26, 2012 #4

    vela

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    Re: Cannot prove the DE is exact??

    Can you show your work in more detail? I'm not sure what you're doing. Your expression for ##\partial F/\partial x## isn't correct, and I'm not sure where you got ##e^{2x}+A'(y)## from.
     
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