variable separable!

**lawochekel** · September 17th 2012, 02:38 AM

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

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!

**kalyanram** · September 17th 2012, 02:45 AM

Try $y=u^2$ substitution.

**Prove It** · September 17th 2012, 02:54 AM

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

Thread: variable separable!

LinkBack

Thread Tools

Display

variable separable!

Re: variable separable!

Re: variable separable!

Similar Math Help Forum Discussions

Variable separable question.

Solving for a variable fraction, multiplied by a non variable.

time derivative of a variable os inversly proportional to other variable

separable extension and separable polynomial

Conversion (Trigonometry Variable To Standard Variable)

Search Tags