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
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
??
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.
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!!!