1. ## differential equations

a sphere with radius of 1meter has a temp = 15degrees. It lies inside a concentric sphere with radius of 2meters and temp = 25degrees. The temp T(r) at a distance r from the common center of the spheres satisfies the differential equation

(d^2T)/dr^2 +(2/r)(dT/dr) = 0

if we let S=dT/dr then S satisfies a first order differential equation. Solve it to find an expression for the temperature T(r) between the spheres.

2. Originally Posted by noscbs
a sphere with radius of 1meter has a temp = 15degrees. It lies inside a concentric sphere with radius of 2meters and temp = 25degrees. The temp T(r) at a distance r from the common center of the spheres satisfies the differential equation

(d^2T)/dr^2 +(2/r)(dT/dr) = 0

if we let S=dT/dr then S satisfies a first order differential equation. Solve it to find an expression for the temperature T(r) between the spheres.
$S = \frac{dT}{dr}$

so

$\frac{dS}{dr} + \frac{2}{r}S = 0$.

Integrating Factor $= e^{\int{\frac{2}{r}\,dr}}$

$= e^{2\ln{r}}$

$= e^{\ln{r^2}}$

$= r^2$.

So multiply through by the Integrating Factor

$r^2\left(\frac{dS}{dr} + \frac{2}{r}S\right) = 0r^2$

$r^2\frac{dS}{dr} + 2rS = 0$

$\frac{d}{dr}(r^2 S) = 0$

$r^2S = \int{0\,dr}$

$r^2S = C_1$

$S = C_1r^{-2}$.

Now remembering that $S = \frac{dT}{dr}$

$\frac{dT}{dr} = C_1r^{-2}$

$T = \int{C_1r^{-2}\,dr}$

$= C_2 - C_1r^{-1}$.

Now use your initial conditions to find $C_1$ and $C_2$.

3. I do not understand where the integrating factor came from, and I also do not understand how you jumped from to
if you would please explain these further that would be great. Also i do not understand why my post was moved out of calculus section, because this is only a calc2 problem, so perhaps this math is higher than my current level

4. nvm sorry if you had to waste your time reading these replies, i understand everything perfectly.

5. If you are having trouble with the interating factor technique, in this case, the first order DE is also seperable.

$\frac{dS}{dr} + \frac{2}{r}S = 0$

$\frac{dS}{dr} = -\frac{2}{r}S$

$\frac{1}{S}\,\frac{dS}{dr} = -\frac{2}{r}$

$\int{\frac{1}{S}\,\frac{dS}{dr}\,dr} = \int{-\frac{2}{r}\,dr}$

$\int{\frac{1}{S}\,dS} = -2\ln{|r|} + C_1$

$\ln{|S|} + C_2 = -\ln{|r^2|} + C_1$

$\ln{|S|} + \ln{|r^2|} = C$, where $C = C_1 - C_2$

$\ln{|Sr^2|} = C$

$|Sr^2| = e^C$

$Sr^2 = \pm e^C$

$Sr^2 = A$, where $A = \pm e^C$

$S = Ar^{-2}$

which is the same answer as given before.