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Homework Help: Help solving the initial value problem

  1. Mar 25, 2012 #1
    Solve the initial value problem 2(√x)y'+y+4(√x) ; y(1)=2

    I am having trouble separating the x's and y's in order to integrate. I keep coming up with:

    dy/dx +y/(2(√x))=2.....

    What do I keep missing here? Im pretty sure you leave the y(1)=2 alone until you are finished with the integration, in which case you plug in x=1 and y=2 to solve for the constant. Is this the correct thought process for this problem?

  2. jcsd
  3. Mar 25, 2012 #2
    Or can I substitute in the y(1) for the 2 on the right side of the equation and make all the other x's equal to 1?
  4. Mar 25, 2012 #3
    let gif.latex?P(x)=\frac{1}{2\sqrt&space;x}.gif

    and gif.gif

    find the integrating factor gif.gif

    and then use gif.gif

    hope that help :)
  5. Mar 25, 2012 #4
    Ok so now I have:


    p(x)=-1/(2√x) q(x)=2




    this is the solution for all values of x with C=0, correct? Or am I forgetting a step?

  6. Mar 25, 2012 #5
    you firstly have to integrate the right side gif.gif before you multiply it by gif.gif !
  7. Mar 25, 2012 #6

    y=-4sqrt(x)-4+C with C=10 is the answer given the IC y(1)=2?
  8. Mar 25, 2012 #7
    when u divide the right side after integrating by gif.gif , u should also divide the constant too so it should be

  9. Mar 25, 2012 #8


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    Do you mean [itex]2\sqrt{x}y'+ y= 4\sqrt{x}[/itex]?

  10. Mar 25, 2012 #9
    yes I do.

  11. Mar 25, 2012 #10
    wouldn't that just be another constant?

    so instead of writing Ce^sqrt(x), you could just consider it a new constant
  12. Mar 25, 2012 #11


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    C is a constant, but [itex]Ce^\sqrt{x}[/itex] is definitely not a constant.
  13. Mar 25, 2012 #12
    isnt that irrelevant since there is an IC? wouldnt the x in that expression =1 and then just equal C multiplied by another constant?

    edit: Are you saying i should divide 10 by e^1 to find the constant?
  14. Mar 26, 2012 #13


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    No, that's not irrelevant.

    In using the Initial Condition to evaluate the constant, C , you do set x=1, but in the overall solution there is a term [itex]Ce^\sqrt{x}[/itex], unless the Initial Condition gives C=0, which is not the case here.
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