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Linear Equation

  1. Jul 29, 2006 #1
    I am new to to this topic, hints?


    The only examples are in the form dy/dx+p[x(y)]=Q[x]
  2. jcsd
  3. Jul 29, 2006 #2


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    Try some simple forms for the solution. y=ax+b should work.

    If you want to be less reliant on luck, you could try looking at the asymptotic behavior of the DE. It seems to resemble the equation for exponential growth, in which case the y term would soon dominate over the x term on the RHS, and the solution would approach true exponential growth more and more closely. This suggests trying a solution of the form y=f(x) e5x. Plugging this in and deriving a DE for f(x), you get something you can solve easily, but it turns out the exponential cancels, and you're left with something of the above form.
  4. Jul 30, 2006 #3


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    Do you mean the differential equation y'= x+ 5y?

    You say "The only examples are in the form dy/dx+p[x(y)]=Q[x]". Surely you must mean dy/dx+ p(y(x))= Q(x). This is exactly of that form:
    dy/dx- 5y= x. p(y)= 5y and Q(x)= x. StatusX's suggestion of trying y= f(x)e5x is excellent but you could do basically the same thing by multiplying the entire equation by e5x (an "integrating factor"):
    [tex]e^{5x}\frac{dy}{dx}+ 5e^{5x}y= xe^{5x}[/tex]
    [tex]\frac{d(e^{5x}y)}{dx}= e^{5x}\frac{dy}{dx}+ 5e^{5x}y[/tex]
    by the product rule. The problem reduces to an integration by parts.
  5. Jul 30, 2006 #4


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    Right, sorry. I've been doing DEs lately where there's no obvious solution, and these are the methods I've used. I forgot that equations of this form have a general method of solution, namely integrating factors. Thanks Halls.
  6. Aug 1, 2006 #5
    soo..[tex] \int \frac{d}{dx}(e^5^xy) - \int 5e^5^x = 5e^5^xy[/tex]?
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