Results 1 to 6 of 6

Thread: help! integration / confusion

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    53

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2009
    Posts
    53
    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
    ??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    5
    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 $\displaystyle yy'=x,$ it follows $\displaystyle (\frac{y^2}{2})'=x.$ So we can let $\displaystyle u=\frac{y^2}{2},$ consequently $\displaystyle u'=x$ which tells us
    $\displaystyle u=\frac{x^2}{2}+C.$ That is, $\displaystyle y^2=x^2+C'.$ To this point, it remains to find out the exact value of $\displaystyle C'.$ Leave the remainder up to you.

    By the way, your way makes no sense.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2009
    Posts
    53
    Quote Originally Posted by mchaelzang View Post
    From $\displaystyle yy'=x,$ it follows $\displaystyle (\frac{y^2}{2})'=x.$ So we can let $\displaystyle u=\frac{y^2}{2},$ consequently $\displaystyle u'=x$ which tells us
    $\displaystyle u=\frac{x^2}{2}+C.$ That is, $\displaystyle y^2=x^2+C'.$ To this point, it remains to find out the exact value of $\displaystyle C'.$ Leave the remainder up to you.

    By the way, your way makes no sense.

    say what?!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    53
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    5
    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 $\displaystyle yy'=x$ can be derived from $\displaystyle \int^{a}_{-a}y^2dx=1.$ Then it means we have two facts that $\displaystyle yy'=x$ and $\displaystyle \int^{a}_{-a}y^2dx=1.$ From my last reply, we get the function $\displaystyle y$ must satisfy $\displaystyle y^2=x^2+C'.$ However, it is wrong. For example, let $\displaystyle y(x)\equiv \sqrt{\frac{1}{2a}}$ for all $\displaystyle x,$ which also satisfies $\displaystyle \int^{a}_{-a}y^2dx=1.$ A contradiction!

    pepsi, actually, you should have known that $\displaystyle \int^{a}_{-a}y^2dx=1$ is an integral equality, while $\displaystyle yy'=x$ is a differential equation. pay attention to theire solution sets of functions, respectively!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. log rule for integration confusion
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jul 15th 2011, 08:51 AM
  2. Integration Confusion
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 22nd 2010, 12:38 AM
  3. Integration Confusion
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Mar 21st 2010, 06:54 PM
  4. Integration by Substitution - Confusion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 23rd 2010, 09:29 AM
  5. Integration Confusion
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 28th 2006, 06:07 PM

Search Tags


/mathhelpforum @mathhelpforum