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When solving differential equations, a first-order differential equation is often encountered. These equations involve derivatives where the highest derivative is of the first order. In our exercise, we start with a second-order differential equation:

• The given equation is \( \frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr} = 0 \).
• By substituting \( S = \frac{dT}{dr} \), this transforms the equation into a first-order form: \( \frac{dS}{dr} + \frac{2}{r}S = 0 \).

First-order differential equations can often be simpler to solve because they involve only one derivative term. In this case, the equation is separable, which is a convenient class of first-order equations. Here, we separate variables to find an integrable form. This step is crucial as it simplifies the process of integration.
Understanding how to form and solve first-order differential equations is essential in solving more complex problems in calculus and physics.

Boundary conditions are specific values that a solution to a differential equation must satisfy. They come into play after you solve the differential equation and find a general solution involving constants.

• In our example, the boundary conditions are \( T(1) = 15 \text{ and } T(2) = 25 \).

We apply these conditions after solving for \( T(r) \) from the differential equation. They are used to find the constants in the temperature equation. This process transforms a general solution into a specific one that fits the physical scenario or problem at hand.
To apply these conditions, you substitute the boundary values into your equation. This results in a system of equations that helps determine the values of the constants. It is a critical step because it anchors your mathematical solution to the real-world problem, ensuring the relevance and accuracy of the model.

Integration is a mathematical tool used to find functions given their derivatives. It plays a crucial role in solving differential equations. In this exercise, several integration techniques are applied:

• The first integration is performed after separating variables \( \frac{dS}{S} = -\frac{2}{r} dr \), leading to an integral form for \( \ln|S| = -2 \ln|r| + C_1 \).
• The second integration happens when solving for \( T(r) \). Here \( S = C_2 r^{-2} \) is integrated to yield the temperature function \( T(r) = -C_2 r^{-1} + C_3 \).

Integration here is performed using common techniques such as variable separation and simple power-rule integration. These techniques allow us to transform a differential equation into a function, presenting a practical solution. Consistency in honing integration skills is vital, as they are broadly applicable across various mathematical and applied problems.