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Differential Equation expression help

  1. Aug 3, 2010 #1
    Here's the question:

    Let f be the function satisfying f'(x)=x[tex]\sqrt{f(x)}[/tex] for all real numbers where f(3)=25.

    a. Find f''(3)

    b. Write an expression for y-f(x0 by solving the differential equation [tex]\frac{dy}{dx}[/tex] = x[tex]\sqrt{y}[/tex] with the initial condition of f(3)=25.

    For a, I got [tex]\frac{x^2}{2}[/tex]+[tex]\sqrt{f(x)}[/tex], so my answer was [tex]\frac{19}{2}[/tex].

    For b, I immediately substituted, getting dy/dx=3sqrt(25). then, dy/dx=15 -> dy=15dx -> integrate, y=15x+c, and since the initial condition is f(3)=25, by substitution, C=-20. My answer in the end was y=15x-20.

    Am I on the right track? Thanks.
  2. jcsd
  3. Aug 3, 2010 #2


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    I haven't actually studied differential equations, so I don't know any methods to solve them, and this is coming from my own intuition so I advise you to read with caution :smile:



    would satisfy this differential.

    I don't quite understand what part b) is asking though. Part a) is pretty easy if you know what the function is.
  4. Aug 3, 2010 #3
    How did you get to that result? It satisfies, but I couldn't reach that result..
  5. Aug 3, 2010 #4

    Gib Z

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    Separation of Variables. Also, your expression for the second derivative was correct, but 19/2 isn't so check your arithmetic.
  6. Aug 3, 2010 #5
    well, 9/2 + 5 is 19/2, or 9.5 isn't it?

    Also, can you describe me the process?

  7. Aug 3, 2010 #6

    Gib Z

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    Ahh foolish me, I put in x=5 rather than x=3, you are correct.

    To separate variables, start with [tex]
    [/tex] and then take over the [itex]\sqrt{y}[/itex] term to give [tex]\frac{1}{\sqrt{y}}\frac{dy}{dx} = x[/tex], then integrate both sides with respect to x.
  8. Aug 3, 2010 #7


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    This part is important to note:
    I just realized that to get the square root of the function and have an x multiplied by it, I need a quadratic of the form ax2+c taking all to the 2nd power since we are going to minus one from the power, which is also the square root.

    So we have [tex]y=(ax^2+c)^2[/tex] and then [tex]y'=2(ax^2+c).2ax=4ax(ax^2+c)[/tex]

    Now we need a=1/4 to satisfy the problem.
  9. Aug 3, 2010 #8


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    Oh and Gib Z's method works much better for those that don't simply notice it! It did take longer to solve, but there, I've learnt differentials.
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