1. ## variable separable!

$dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)$

have reduce to $\int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx$

i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

thanks!

2. ## Re: variable separable!

Try $y=u^2$ substitution.

3. ## Re: variable separable!

Originally Posted by lawochekel
$dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)$

have reduce to $\int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx$

i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

thanks!
\displaystyle \begin{align*} \int{\frac{1}{2\sqrt{y} - 2y}\,dy} &= \int{\frac{1}{2\sqrt{y}\left( 1 - \sqrt{y} \right)}\,dy} \\ &= -\int{\frac{1}{1 - \sqrt{y}}\left( -\frac{1}{2\sqrt{y}}\right) dy} \end{align*}

Now make the substitution \displaystyle \begin{align*} u = 1 - \sqrt{y} \implies du = -\frac{1}{2\sqrt{y}}\,dy \end{align*} and the integral becomes \displaystyle \begin{align*} -\int{\frac{1}{u}\,du} \end{align*}. I'm sure you can go from here