differential equations

noscbs · #1 $Old$ November 1st, 2009, 11:02 PM

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.

Prove It · #2 $Old$ November 1st, 2009, 11:17 PM

Quote:

Originally Posted by noscbs $View Post$

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 $C_2$ .

noscbs · #3 $Old$ November 2nd, 2009, 06:10 PM

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

noscbs · #4 $Old$ November 2nd, 2009, 06:26 PM

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

Prove It · #5 $Old$ November 2nd, 2009, 06:40 PM

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.