A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as `k= a//T`, where a is a constant. The ends of the rod are kept at temperatures `T_1 and T_2`. Find the function T(x), where x is the distance from the end whose temperature is `T_1`.

Explanation

To solve the problem, we need to find the temperature distribution \( T(x) \) along the length of a rod with varying thermal conductivity. The thermal conductivity \( k \) is given by \( k = \frac{a}{T} \), where \( a \) is a constant, and the ends of the rod are maintained at temperatures \( T_1 \) and \( T_2 \).

### Step-by-Step Solution:

1. **Understanding the Heat Flow**:
- The rod has a length \( L \) and is insulated on the sides, meaning heat can only flow along its length. At steady state, the heat current \( \frac{dQ}{dt} \) is constant.

2. **Setting Up the Heat Current Equation**:
- Consider an elemental length \( dx \) at a distance \( x \) from the end at temperature \( T_1 \). The temperature at this point is \( T \), and at the other end of the element, it is \( T + dT \).
- The heat current through this elemental length can be expressed as:
\[ \frac{dQ}{dt} = k A \frac{dT}{dx} \]
- Here, \( A \) is the cross-sectional area of the rod.

3. **Substituting for k**:
- Since \( k = \frac{a}{T} \), we can substitute this into the heat current equation:
\[ \frac{dQ}{dt} = \frac{a}{T} A \frac{dT}{dx} \]

4. **Setting the Heat Current Equal**:
- Since the heat current is constant, we can denote it as \( I \):
\[ I = \frac{a}{T} A \frac{dT}{dx} \]

5. **Rearranging the Equation**:
- Rearranging gives us:
\[ I \cdot T = a A \frac{dT}{dx} \]
- Rearranging further, we have:
\[ \frac{dT}{dx} = \frac{I T}{a A} \]

6. **Integrating the Equation**:
- To find \( T \) as a function of \( x \), we can separate variables and integrate:
\[ \int \frac{dT}{T} = \frac{I}{a A} \int dx \]
- This leads to:
\[ \ln T = \frac{I}{a A} x + C \]
- Where \( C \) is the integration constant.

7. **Finding the Constant of Integration**:
- We can express \( C \) using the boundary condition at \( x = 0 \) where \( T = T_1 \):
\[ \ln T_1 = C \]
- Thus, we have:
\[ \ln T = \frac{I}{a A} x + \ln T_1 \]

8. **Exponentiating to Solve for T**:
- Exponentiating both sides gives:
\[ T = T_1 e^{\frac{I}{a A} x} \]

9. **Using the Boundary Condition at x = L**:
- At \( x = L \), \( T = T_2 \):
\[ T_2 = T_1 e^{\frac{I}{a A} L} \]
- From this, we can solve for \( I \):
\[ \frac{I}{a A} = \frac{\ln \frac{T_2}{T_1}}{L} \]

10. **Final Expression for T(x)**:
- Substituting back into the equation for \( T \):
\[ T(x) = T_1 e^{\frac{\ln \frac{T_2}{T_1}}{L} x} \]
- This simplifies to:
\[ T(x) = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}} \]

### Final Answer:
\[ T(x) = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}} \]