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Linear differential equation

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    equation (1) y' + (ylny)/x = xy

    we set y(x)= eu(x)
    Equation (1 ) proposed becomes an equation
    1st order linear differential type : equation (2) u ' + p( x) u = q ( x)

    a) Find the equation ( 2) using the change of variable proposed above.

    2. Relevant equations


    3. The attempt at a solution
    equation (1) y' + (ylny)/x = xy
    y' + (eu(x)ln eu(x))/x = x*eu(x)

    y' + (eu(x)ln eu(x)) = x2*eu(x)

    pretty much after that, I have no idea how to proceed, please help :)
     
  2. jcsd
  3. Oct 3, 2015 #2

    andrewkirk

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    I think they want you to say what p(x) and q(x) are.

    Your second line in section 3 is correct, but your 3rd is not. Go back to the second line and continue substituting ##e^{u(x)}## for ##y##, by differentiating it wrt ##x## and replacing y' by that. Then do what's necessary to make that equation have the same form as (2).
     
  4. Oct 3, 2015 #3
    eu(x)' + (eu(x))ln eu(x))/x = x*eu(x)
    (eu(x))' + ((u(x))*eu(x))/x = x*eu(x)

    Like that ?
     
    Last edited: Oct 3, 2015
  5. Oct 3, 2015 #4

    andrewkirk

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    No. The division by x in the 2nd term on the LHS has disappeared. Why? If you want to multiply everything by x, you need to multiply all other terms by x.

    Also, what do you mean by eu(x)'? What you have written is ambiguous. If you mean e[u(x)'] then that's incorrect. Alternatively, if you mean [eu(x)]' then that's correct, but you need to take the next step of performing the differentiation (wrt x).
     
  6. Oct 3, 2015 #5
    I am not sure how to do the differentiation (wrt x).

    and I fixed the mistake in my last reply, sorry.
     
  7. Oct 3, 2015 #6

    andrewkirk

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    The first term is ##(e^{u(x)})'## which means ##\frac{d}{dx}e^{u(x)}##. To perform the differentiation, use the chain rule.
     
  8. Oct 3, 2015 #7
    I need to use the chain rule only for (d/x)*eu(x)?
     
  9. Oct 3, 2015 #8

    andrewkirk

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    Where else would you use it? That's the only differentiation you need to do, because that's the only prime (') in the equation. Prime after a term means differentiate that term.
     
  10. Oct 3, 2015 #9
    it gives me that
    eu(x)*(d/x)*(u(x))
     
  11. Oct 3, 2015 #10
    eu(x)*(d/x)*(u(x)) + ((u(x)*eu(x)))/x = x*eu(x)

    (d/x)*(u(x)) + (((u(x)*eu(x)))/x )/eu(x)= (x*eu(x))/eu(x)

    (d/x)*(u(x)) + u(x)*/x= x

    P(x) = 1/x
    Q(x)=x
     
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