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Please check if the differential equation is correct

  1. Feb 5, 2012 #1
    I am practising for my test. The question is to solve a differential equation

    dy/dx + y/x + 1 = 5x

    y(0) = 1.

    The answer that i have come up with is

    (xy+y)= 5x^3/3+5x^2/2+c

    by substituting the values x=0 and y=1 in to the general equation I get

    y(x+1)=5x^3/3 + 5x^2/2 +1

    as the particular solution.

    Can you tell me how will the particular solution look like and why this particular solution exists?
  2. jcsd
  3. Feb 5, 2012 #2


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    Science Advisor

    Hey rohit99 and welcome to the forums.

    For your equation your initial condition only effects a constant in the entire equation.

    Based on this what is the effect of adding or subtracting a constant in a general equation? (Hint: how does it 'shift' the function?)
  4. Feb 5, 2012 #3


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    Science Advisor

    You mean y/(x+1).

    What exactly do you mean by "look like"? If you just mean "solve for y", divide both sides by x+ 1.
    The differential equation can be written
    [tex]\frac{dy}{dx}= 5x- \frac{y}{x+1}[/tex]
    The function on the right side is differentiable for all y and all x except -1 so by the "fundamental existence and uniqueness theorem" for initial value problems, a unique solution to this problem exist for all x larger than -1.
    Last edited by a moderator: Feb 6, 2012
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