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Math Help - Verify that the indicated function is a solution to the given differential equation

  1. #1
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    Verify that the indicated function is a solution to the given differential equation

    xy'-sin(x)=0, y = \int_{0}^{x}\frac{sin(t)}{T}dt


    So this is what I did. I integrated the function to get, \frac{1}{T}[-cos(x)+cos(1)]
    Then took the derivative of that with respect to x to get \frac{sin(x)}{T}

    Then I sub it in to get
    \frac{xsin(x)}{T} - sin(x) which does not equal 0, so i'm saying that this function is not a solution to the differential equation.

    IS this correct?
    David.
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  2. #2
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    Re: Verify that the indicated function is a solution to the given differential equati

    I see no reason to integrate when you just turn around and differentiate. With y= \int_0^x\frac{sin(t)}{T}dt by the "Fundamental Theorem of Calculus", that the derivative of \int_a^x f(t) dx is just f(x), we get immediately your y'= \frac{sin(t)}{T}. And that does NOT satisfy the differential equation, again as you say.

    HOWEVER, I think you have misread the problem. If y= \int_0^x \frac{sin(t)}{t}dt, so that the denominator is the variable, not a constant, then the function becomes very hard to integrate but it is easy to use the "Fundamental Theorem of Calculus" to get y'= \frac{sin(x)}{x}. And that clearly does satisfy the equation.

    (By the way, the constant in your integral should be cos(0)= 1, not cos(0)- but that was probably a typo.)
    Last edited by HallsofIvy; July 27th 2012 at 04:53 PM.
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  3. #3
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    Re: Verify that the indicated function is a solution to the given differential equati

    Yeah sorry about that. The typo was actually in the question, not my answer. The limits of integration was from 1 - x. So it should have been a 1 at the bottom end of the integral sign. ANd the question had a capital T for the denominator as I wrote. But I think you're right and actually the capital T should have been a small t, because they give a hint using the fundamental theorum of calculus just like you stated when you were talking about it should have been a small t.

    Yeah, so I will assume they meant a small t instead of a big T and use the Fundamental THeorem of Calculus as you did.

    Thanks for your help. I don't know if the 1 makes that harder now, but maybe that was a typo in the book too.
    Thanks.
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