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Math Help - help! integration / confusion

  1. #1
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    help! integration / confusion

    can anyone see some way of getting to yy'=x
    from
    integral of y^2 from -a to a (dx) = 1 ?

    p.s. i am trying d/dx {integral of y^2 from -a to a (dx)} =0, but find my self lost!

    thanks
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  2. #2
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    does the following make sense?

    d/dx {integral of y^2 from -a to a (dx)} =0
    =>
    integral of d(y^2)/dx from -a to a (dx) =0
    =>
    integral of 2yy' from -a to a (dx) =0
    =>
    integral of yy' from -a to a (dx) =0
    =>
    integral of yy' from -a to a (dx) =integral of x from -a to a = 0
    =>
    yy'=x
    ??
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  3. #3
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    Quote Originally Posted by pepsi View Post
    can anyone see some way of getting to yy'=x
    from
    integral of y^2 from -a to a (dx) = 1 ?

    p.s. i am trying d/dx {integral of y^2 from -a to a (dx)} =0, but find my self lost!

    thanks
    From yy'=x, it follows (\frac{y^2}{2})'=x. So we can let u=\frac{y^2}{2}, consequently u'=x which tells us
    u=\frac{x^2}{2}+C. That is, y^2=x^2+C'. To this point, it remains to find out the exact value of C'. Leave the remainder up to you.

    By the way, your way makes no sense.
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  4. #4
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    Quote Originally Posted by mchaelzang View Post
    From yy'=x, it follows (\frac{y^2}{2})'=x. So we can let u=\frac{y^2}{2}, consequently u'=x which tells us
    u=\frac{x^2}{2}+C. That is, y^2=x^2+C'. To this point, it remains to find out the exact value of C'. Leave the remainder up to you.

    By the way, your way makes no sense.

    say what?!!
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  5. #5
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    Quote Originally Posted by pepsi View Post
    say what?!!
    right, sorry. let me start again. if i gave the impression that i was looking for a function y(x) such that yy'=x i apologise. i meant to find out if i can deduce (not assume) that yy'=x from integral of y^2 from -a to a (dx) =1
    pepsi
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  6. #6
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    Quote Originally Posted by pepsi View Post
    right, sorry. let me start again. if i gave the impression that i was looking for a function y(x) such that yy'=x i apologise. i meant to find out if i can deduce (not assume) that yy'=x from integral of y^2 from -a to a (dx) =1
    pepsi
    okay, now suppose that yy'=x can be derived from \int^{a}_{-a}y^2dx=1. Then it means we have two facts that yy'=x and \int^{a}_{-a}y^2dx=1. From my last reply, we get the function y must satisfy y^2=x^2+C'. However, it is wrong. For example, let y(x)\equiv \sqrt{\frac{1}{2a}} for all x, which also satisfies \int^{a}_{-a}y^2dx=1. A contradiction!

    pepsi, actually, you should have known that \int^{a}_{-a}y^2dx=1 is an integral equality, while yy'=x is a differential equation. pay attention to theire solution sets of functions, respectively!!!
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