The question is

$\displaystyle \frac{du}{dr} = \frac {1+\sqrt{r}}{1+\sqrt{u}} $

And I am supposed to solve for u.

The answer I got was

$\displaystyle u + \frac{2\sqrt{u^3}}{3} = r + \frac{2\sqrt{r^3}}{3} + C$

Is there a way I can solve for u?

Printable View

- Apr 9th 2009, 11:11 AMxwanderingpoetxSeperable Differential EQ
The question is

$\displaystyle \frac{du}{dr} = \frac {1+\sqrt{r}}{1+\sqrt{u}} $

And I am supposed to solve for u.

The answer I got was

$\displaystyle u + \frac{2\sqrt{u^3}}{3} = r + \frac{2\sqrt{r^3}}{3} + C$

Is there a way I can solve for u? - Apr 9th 2009, 11:29 AMMoo
Hello,

Is there an initial condition so that you can get C ?

Because if C is 0, I have this factorisation :

$\displaystyle (\sqrt{u}-\sqrt{r})\left(\frac 23 \cdot u+\frac 23 \sqrt{ur}+\frac 23 \cdot r+\sqrt{u}+\sqrt{r}\right)=0$

Since you talked about $\displaystyle \sqrt{r}$ and $\displaystyle \sqrt{u}$, we can assume that u and r are positive.

Thus the solutions are : u=r, or u=r=0 (for the second term, because it's always $\displaystyle \geq 0$) - Apr 9th 2009, 11:34 AMxwanderingpoetx
$\displaystyle C$ is just the arbitrary constant associated with the integration. But it isn't given a value so it needs to be in the final equation.