10.4. Examples of Direct Integration

Example 1: ∂𝑧 = 2𝑥 +𝑦 ∂𝑥 Solution: Integrate with respect to 𝑥: 𝑧 = ∫(2𝑥+ 𝑦)𝑑𝑥 = 𝑥² + 𝑥𝑦 + 𝜙(𝑦) 𝜙(𝑦) is an arbitrary function of 𝑦.

In this example, we're given a first-order PDE: the partial derivative of z with respect to x equals 2x + y. To solve it using direct integration, we perform the following steps:

1. We treat y as a constant and integrate the function (2x + y) with respect to x.
2. The integral of 2x is x², and the integral of y is xy.
3. After integration, we add an arbitrary function of y, denoted as 𝜙(y), since we are integrating with respect to x and y acts as a constant. This process helps us find a general solution for z in terms of x and y.

Imagine you're determining the height of water in a tank as it fills up over time. The rate of height increase (∂z/∂x) depends on how fast water flows in (2x + y). By integrating this rate, you're effectively figuring out the total height of water (z) after a certain period, while accounting for other variables like the constant flow from a garden hose (which is represented by y).

Example 2: ∂𝑧 = 𝑥²𝑦 +𝑦³ ∂𝑦 Solution: Integrate with respect to 𝑦: 𝑧 = ∫(𝑥²𝑦 + 𝑦³)𝑑𝑦 = 𝑥²𝑦² + rac{1}{4}𝑦⁴ + 𝜓(𝑥) 𝜓(𝑥) is an arbitrary function of 𝑥.

Here, we're given a different first-order PDE: the partial derivative of z with respect to y equals x²y + y³. The steps to solve it are as follows:

1. Treat x as a constant and integrate the function (x²y + y³) with respect to y.
2. The integral of x²y is (1/2)x²y², and the integral of y³ is (1/4)y⁴.
3. We combine these results to obtain the expression for z and add an arbitrary function of x, denoted as 𝜓(x). This step accounts for the fact that z could vary with x independently of y.

Think of a farmer who is trying to determine the yield (z) of crops based on the amount of sunlight and rain. The yield depends on both sunlight (x²y) and other factors like rainfall (y³). As the farmer integrates the effects of rainfall over the growing season, they find the overall yield while considering the stable conditions (represented by the arbitrary function 𝜓(x)) that affect the crops.

Example 3 (Two partials given): ∂𝑧 ∂𝑧 = 𝑥+ 𝑦, = 𝑥− 𝑦 ∂𝑥 ∂𝑦 Step 1: Integrate the first equation: 𝑧 = rac{𝑥²}{2} + 𝑥𝑦 + 𝜙(𝑦) Step 2: Differentiate this with respect to 𝑦: ∂𝑧 = 𝑥 + 𝜙′(𝑦) ∂𝑦 Compare with the second equation: 𝑦² 𝑥 + 𝜙′(𝑦) = 𝑥 − 𝑦 ⇒ 𝜙′(𝑦) = −𝑦 ⇒ 𝜙(𝑦) = −rac{𝑦²}{2} Final Answer: 𝑧 = rac{𝑥²}{2} + 𝑥𝑦 − rac{𝑦²}{2}

In this example, we’re given two equations involving partial derivatives of z, indicating that z is connected to both x and y:

1. From the first equation, we integrate with respect to x while treating y as constant, resulting in an expression for z that includes an arbitrary function of y, 𝜙(y).
2. Next, we differentiate this expression with respect to y to find the corresponding equation, which gives us a new relation involving 𝜙′(y).
3. We compare this to the second partial derivative equation to find 𝜙(y). After solving for 𝜙(y), we substitute it back into our expression for z to find the complete solution, allowing us to understand how z behaves under the influence of both x and y.

Consider a business analyzing profits based on varying production rates (x) and market demand (y). The profit equation depends on both production and demand. By integrating profit margins for different levels of production, then analyzing how changes in market demand affect overall profits lead to a complete understanding of not just how production impacts profits, but also how fluctuations in demand interconnect with production to ultimately shape the business's profitability.

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