5.5. Solved Examples

Solve: ∂𝑧 ∂𝑧 𝑝 + 𝑞 = 𝑧 ⇒ + = 𝑧 ∂𝑥 ∂𝑦 Solution: Here, • 𝑃 = 1, 𝑄 = 1, 𝑅 = 𝑧 Auxiliary equations: 𝑑𝑥 𝑑𝑦 𝑑𝑧 = = 1 1 𝑧 𝑑𝑥 𝑑𝑦 From = ⇒ 𝑥 − 𝑦 = 𝑐_1 1 1 From 𝑑𝑥 = 𝑑𝑧 ⇒ ∫𝑑𝑥 = ∫ 𝑑𝑧 ⇒ 𝑥 = ln𝑧 +𝑐_2 ⇒ 𝑧 = 𝑒^{𝑥−𝑐_2} 1 𝑧 𝑧 2 Now, • 𝑢 = 𝑥 − 𝑦 = 𝑐_1 • 𝑣 = 𝑧𝑒^{−𝑥} = 𝑐_2 General solution: 𝜙(𝑥−𝑦,𝑧𝑒^{−𝑥}) = 0 or 𝑧 = 𝑒^{𝑥}𝑓(𝑥− 𝑦)

In this example, we start with a partial differential equation (PDE) where the relationship between changes in z with respect to x and y is defined. We set P, Q, and R from the equation, identifying them as P=1, Q=1, and R=z. Next, we form the auxiliary equations and solve them. From the equation dx/dy = 1, we find that x - y = c1. We also solve dx/dz, leading to an integration that gives us the relation between x and z. Finally, we identify two independent solutions and express the general solution in terms of u and v, yielding the general solution format.

Imagine you are tracking temperature changes in a room based on time and external weather. If you know the temperature at two different times and can represent those changes with a formula, just as we did with z in the example, you can extrapolate what the temperature will be at any point in the future. This is similar to how we derive solutions in mathematics by looking at the relationships and changes defined in the problem.

Solve: ∂𝑧 ∂𝑧 𝑦𝑝 −𝑥𝑞 = 0 (i.e., 𝑦 −𝑥 = 0) ∂𝑥 ∂𝑦 Solution: Here, • 𝑃 = 𝑦, 𝑄 = −𝑥, 𝑅 = 0 Auxiliary equations: 𝑑𝑥 𝑑𝑦 𝑑𝑧 = = 𝑦 −𝑥 0 From 𝑑𝑥 = 𝑑𝑦 ⇒ 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 ⇒ 𝑑(𝑥^2 + 𝑦^2) = 0 ⇒ 𝑥^2 + 𝑦^2 = 𝑐 𝑦 −𝑥 1 𝑑𝑥 𝑑𝑧 From = ⇒ 𝑑𝑧 = 0 ⇒ 𝑧 = 𝑐_2 𝑦 0 2 General solution: 𝜙(𝑥^2 + 𝑦^2, 𝑧) = 0 or 𝑧 = 𝑓(𝑥^2 + 𝑦^2)

This example addresses a different kind of PDE. We again define P, Q, and R, with distinct values. The auxiliary equations lead us through a series of relationships, illustrating a system where we analyze the change in z by considering a circular symmetry in the x-y plane. We derive that changes in x and y are related to their squares, leading us to the conclusion that they form a circular equation. The final general solution represents a function of the radius squared, indicating that z remains constant along circles defined by the distance from the origin.

Picture a forest as a circular area of trees surrounding a small pond. The further you stand from the center (the pond), the less likely you are to feel a breeze (represented by z in our example). As you walk in a circle around the pond, your position is determined by your distance from the center, similar to how we derived the circular symmetry in the second example. This analogy helps students visualize how variables related to position can affect outcomes in a mathematical sense.

--