Question

A sphere with radius \(1m\)has temperature \(15^\circ C\). It lies inside a concentric sphere with radius\(2m\)and temperature \(25^\circ C\). The temperature \(T\left( r \right)\) at a distance \(r\) from the common center of the spheres satisfies the differential equation

\(\frac{{{d^2}T}}{{d{r^2}}} + \frac{2}{r}\frac{{dT}}{{dr}} = 0\)

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

Short answer

The required function \(T\left( r \right) = - \frac{{20}}{r} + 35\).

Step-by-step solution

Given

Radius of the sphere whose temperature is \(15^\circ C = 1m\).

Radius of the sphere whose temperature is\(25^\circ C = 2m\).

The differential equation is\(\frac{{{d^2}T}}{{d{r^2}}} + \frac{2}{r}\frac{{dT}}{{dr}} = 0\).

Here, \(S = \frac{{dT}}{{dr}}\) be first order differential equation.

Calculation

Consider the differential equation, \(\frac{{dS}}{{dr}} + \frac{2}{r}S = 0\).

Separate the variables,

\(\begin{aligned}{}\frac{{dS}}{{dr}} = - \frac{2}{r}S\\\frac{{dS}}{{2S}} = - \frac{{dr}}{r}\end{aligned}\)

Integrate on both sides,

\(\begin{aligned}{}\int {\frac{{dS}}{{2S}} = - \int {\frac{{dr}}{r}} } \\\frac{1}{2}\left( {\ln S} \right) = - \ln r + C\\\left( {\ln S} \right) = - 2\ln r + C\end{aligned}\)

Taking exponential on both sides,

\(\begin{aligned}{}S = {e^{ - 2\ln r + C}}\\ = {e^{\ln {r^{ - 2}}}}.{e^C}\\ = {r^{ - 2}}C\end{aligned}\)

Here,\(C\)is constant.

Since\(S = \frac{{dT}}{{dr}}\),\(dT\)is obtained.

\(\begin{aligned}{}\frac{{dT}}{{dr}} = \frac{C}{{{r^2}}}\\dT = \frac{{dr}}{{{r^2}}}C\end{aligned}\)

Integrate on both sides,

\(\begin{aligned}{}\int {dT = C\int {\frac{{dr}}{{{r^2}}}} } \\T = C\frac{{{{\left( r \right)}^{ - 2 + 1}}}}{{ - 2 + 1}} + k\\T = - \frac{C}{r} + k\end{aligned}\)

Here\(C\)and\(K\)are constant.

\(T\left( r \right) = - \frac{C}{r} + K\)-----(a)

Substitute the condition\({\rm{T}}\left( 1 \right) = 15\)and\(T\left( 2 \right) = 25\)in equation(a),

\(\begin{aligned}{} - \frac{C}{1} + K = 15\\ - C + K = 15\end{aligned}\)-----(b)

\(\begin{aligned}{} - \frac{C}{2} + K = 25\\ - C + 2K = 50\end{aligned}\)-----(c)

Subtract equation(b) and equation(c),

\(\begin{aligned}{} - C + 2K + C - K = 50 - 15\\K = 35\end{aligned}\)

Substitute value in equation(b),

\(\begin{aligned}{} - C + 35 = 15\\ - C = 15 - 35\\ - C = - 20\\C = 20\end{aligned}\)

Substitute the respective of\(K\)and\(C\)in equation(a),

\(T\left( r \right) = - \frac{{20}}{r} + 35\)

Conclusion

Therefore, the required function \(T\left( r \right) = - \frac{{20}}{r} + 35\).