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Y''*y' = x(x+1)

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\frac{d^2 y}{dx^2}\cdot\frac{dy}{dx}=x(x+1), \hspace{10pt} y(0)=1, \hspace{5pt} y'(0)=2[/itex]


    2. Relevant equations

    None I can think of...

    3. The attempt at a solution


    The only thing I even thought to try was turn it into the form:
    [itex]\frac{d^2 y}{dx^2}{dy}=x(x+1){dx}[/itex]
    ...and then it looks kind of like a separable equation?

    I'm lost.
     
  2. jcsd
  3. Feb 27, 2013 #2
    Re: y''*y'=x(x+1)

    Try setting [itex]w=dy/dx[/itex]
     
  4. Feb 27, 2013 #3
    Re: y''*y'=x(x+1)

    That simplifies the equation to:
    [itex]{w}\frac{dw}{dx}=x(x+1)[/itex]

    Integrating both sides with respect to x gives:

    [itex]\int{w}\frac{dw}{dx}dx=\int{x(x+1)}dx[/itex]
    [itex]\frac{w^2}{2}=\frac{(\frac{dy}{dx})^2}{2}=\frac{x^3}{3}+\frac{x^2}{2}+C[/itex]

    Now what? Am I on the right track?
     
    Last edited: Feb 27, 2013
  5. Feb 27, 2013 #4
    Re: y''*y'=x(x+1)

    [tex]wdw=(x^2+x)dx[/tex]
    From here, you merely have to integrate both sides. You can utilize your initial conditions to determine a value for the constant of integration.


    Edit: Yes, you are on the right track. Solve for C, then I am assuming you want to solve for y, however, you never really stated what is being asked.
     
  6. Feb 27, 2013 #5
    Re: y''*y'=x(x+1)

    How can we use y(0)=1? Do we say:
    [itex]\int_{x_o}^{0}wdx+C_2=1[/itex]?
     
  7. Feb 27, 2013 #6
    Re: y''*y'=x(x+1)

    You don't need to use that condition. [itex][[y'(0)]^2/2=[x]^3/3+[x]^2/2+c_1[/itex] since [itex]y'=w[/itex]. Make sense? Can you take it from here?
     
    Last edited: Feb 27, 2013
  8. Feb 27, 2013 #7
    Re: y''*y'=x(x+1)

    I don't understand why you said x(0)? I thought x was the independent variable? So why wouldn't it be:
    [itex]\frac{y'(0)^2}{2}=\frac{2^2}{2}=2=\frac{0^3}{3}+ \frac{0^2}{2}+c_1[/itex]

    ... which means c1=2?

    But I still don't get how that helps me find y?

    that would just change the equation we want to solve to:

    [itex]\frac{{[\frac{dy}{dx}]^2}}{2}=\frac{x^3}{3}+\frac{x^2}{2}+2[/itex]

    But I still don't see how to find y from that. :(
     
    Last edited: Feb 27, 2013
  9. Feb 27, 2013 #8
    Re: y''*y'=x(x+1)

    You are right, that was a typo on my part. Get [itex]dy/dx[/itex] by itself. What is the relationship between y and y'?
     
  10. Feb 27, 2013 #9
    Re: y''*y'=x(x+1)

    So multiply both sides by 2 and then take the square root? Getting:

    [itex]\frac{dy}{dx}=\sqrt{\frac{2x^3}{3}+ x^2+4}[/itex]

    And then integrate with respect to x...

    I don't know how to do that integral, and wolframalpha couldn't do it either. (http://www.wolframalpha.com/input/?i=integrate+sqrt%282x^3%2F3%2Bx^2%2B4%29+with+respect+to+x)

    Sorry to drag this out so long but I just don't get it!
     
  11. Feb 27, 2013 #10
    Re: y''*y'=x(x+1)

    You don't have to necessarily evaluate the integral. You can change the integral to a definite integral; there will be an upper limit and a lower limit. You will change the variable of the integrand from x to some other variable. I can't integrate that either, so that is what I would do. If I missed something, I'm sure someone will jump in.

    ie. [tex]\int{}\frac{e^{at^2}}{t^2}dt=\int_{s_0}^{s}\frac{e^{as^2}}{s^2}ds[/tex]
     
    Last edited: Feb 27, 2013
  12. Feb 27, 2013 #11
    Re: y''*y'=x(x+1)

    [itex]y(x)=\int_{x_o}^{x}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2[/itex]
    So
    [itex]y(0)=\int_{x_o}^{0}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2 = 1[/itex]
    Choose xo=0, so
    [itex]y(0)=\int_{0}^{0}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2 = 0+C_2=1[/itex]
    So C2=1, So the final answer is...

    [itex]y(x)=\int_{0}^{x}\sqrt{\frac{2u^3}{3}+ u^2+4}du+1[/itex]



    Is that right?
     
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