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Easy algebra question related to Differential equations

  1. Sep 11, 2008 #1
    I have this really simple differential equation. It needs to be in the form y'+ p(x)y=q(x) to solve it as a linear first order ODE. I am almost embarassed to ask, but I can't seem to get it in exactly this form. Read below.

    1. The problem statement, all variables and given/known data


    Need y'+ p(x)y=q(x)

    3. The attempt at a solution

    Divide both sides by 2x, gives

    y'/2x + 3y/2x = e^-3x

    Now here I have y' multiplied by the function 1/2x, but according to the proper linear first order ODE form, Y' needs to be by itself.

    Can you give me some tips on how to achieve this? I don't know why I can't see it, its probably very simple, but whatever the algebraic manipulation is for this, I seem to have forgotten it.

    Thank you.
  2. jcsd
  3. Sep 11, 2008 #2


    User Avatar
    Homework Helper

    It's already in that required form. You have y' + 3y =2xe^(-3x). RHS is a function of x, and 3 is p(x). Remember that a constant function is a function of any variable.
  4. Sep 11, 2008 #3
    Ah so p(x)= 3. Yes I see. :wink: Thanks.
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