help! integration / confusion

**pepsi** · July 23rd 2010, 08:10 AM

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

**pepsi** · July 23rd 2010, 08:17 AM

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
??

**mchaelzang** · July 23rd 2010, 09:06 AM

Originally Posted by pepsi

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.

**pepsi** · July 23rd 2010, 10:19 AM

Originally Posted by mchaelzang

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?!!

**pepsi** · July 23rd 2010, 11:36 AM

Originally Posted by pepsi

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

**mchaelzang** · July 23rd 2010, 10:50 PM

Originally Posted by pepsi

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!!!

Thread: help! integration / confusion

LinkBack

Thread Tools

Display

help! integration / confusion

Similar Math Help Forum Discussions

log rule for integration confusion

Integration Confusion

Integration Confusion

Integration by Substitution - Confusion

Integration Confusion

Search Tags