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Solve the initial value problem?

  1. Aug 9, 2010 #1
    1) (dy/dx) = (1) / ( 2√x ) + 10 , y(1) = -1
    2) (d^(2)r / dt^2) = ( 4 / t^3 ) ; (dr/dt) | t=1 = -5, r(1)=5

    ... i have no idea where to start.
    what are the steps needed to figure out the answer...can someone please solve this for me! t
    thank you so much.
     
  2. jcsd
  3. Aug 9, 2010 #2

    quasar987

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    (1) So the problem here is that you need to find a function y(x) such that when you differentiate it, you get (1) / ( 2√x ) + 10, right? Recall that the operation of differentiating is linear. Meaning that for any two functions f,g, we have (f + g)'=f ' + g'. This observation may help you to simplify the problem. Indeed, if you can find two functions f(x), g(x) such that f '(x)=(1) / ( 2√x ) and g'(x)=10, then the solution to the problem will be y(x) = f(x) + g(x). Do you agree?

    Ok, so we've split the difficult problem of finding a function y(x) such that y'(x) = (1) / ( 2√x ) + 10 into the 2 easier problems of finding functions f(x), g(x) such that f '(x)=(1) / ( 2√x ) and g'(x)=10.

    This is just a matter of "differentiating backwards". You are used to finding the derivative of a given function. Here, you are given a function and must find it is the derivative of what function.

    If you have practiced differentiating enough, then this should not be too hard. Start with finding g(x) as it is somewhat easier than finding f(x). Help yourself with a table listing the rules of differentiation if necessary.

    Equivalently, if you know what an integral is and if you're familiar with the rules for solving them, then

    [tex]f(x)=\int \frac{1}{2\sqrt{x}}dx[/tex]

    and

    [tex]g(x)=\int 10dx[/tex]
     
    Last edited: Aug 9, 2010
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