Question

A sphere with radius 1 m has temperature 10°C. It lies inside a concentric sphere with radius 2 m and temperature 28°C. The temperature T(r) (in °C) at a distance r (in meters) from the common center of the spheres satisfies the differential equation d2T dr2 + 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. (Use T for T(r).)

Answer 1

Answer:T=$\frac{-26}{r}$+36

Explanation:

Given Temperature at r=1m is $10^{\circ}C$

Temperature at r=2m is $28^{\circ}C$

$\frac{\mathrm{d^2} T}{\mathrm{d} r^2}+\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=0$

Let $\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=S$

$\frac{\mathrm{d^2} T}{\mathrm{d} r^2}=\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}$

therefore $\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}+\frac{2}{r}$S=0

$\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}=-\frac{-2S}{r}$

solving

$r^2$ S=constant

substitute S value

$\frac{\mathrm{d}T}{\mathrm{d}r}$=$\frac{c}{r^2}$

Solving it we get

T=$\frac{-c}{r}+c_2$

Now using given condition

10=$\frac{-c}{1}+c_2$

28=$\frac{-c}{2}+c_2$

$c_2$=36,c=26

putting c values

T=$\frac{-26}{r}$+36