Problem 9 - Thomas Calculus in SI Units

Question

Solve the differential equations. $$2 \sqrt{x y} \frac{d y}{d x}=1, \quad x, y>0$$

Step-by-Step Solution

Answer

The solution is \(y = \left(\frac{3}{2}(\sqrt{x}+ C)\right)^{2/3}\), where \(C\) is a constant of integration.

Step 1: Separate the Variables

Starting with the given differential equation \(2 \sqrt{x y} \frac{d y}{d x} = 1\), our goal is to separate the variables. We can write this equation as: \(2 \sqrt{x y} \cdot \frac{d y}{d x} = 1\) or equivalently \(\frac{d y}{d x} = \frac{1}{2\sqrt{x y}}\). Multiply through by \(dx\) and divide by \(\sqrt{y}\) to isolate \(dy\) on one side: \(\sqrt{y} \cdot dy = \frac{1}{2\sqrt{x}} \cdot dx\).

Step 2: Integrate Both Sides

Now that we have separated the variables, integrate both sides. The left side becomes \(\int \sqrt{y} \, dy\), which is \(\frac{2}{3} y^{3/2} + C_1\). The right side becomes \(\int \frac{1}{2\sqrt{x}} \, dx\), which is \(\sqrt{x} + C_2\). Thus, we have: \(\frac{2}{3} y^{3/2} = \sqrt{x} + C\), where \(C = C_2 - C_1\).

Step 3: Solve for y

To solve for \(y\), multiply both sides by \(\frac{3}{2}\): \(y^{3/2} = \frac{3}{2} (\sqrt{x} + C)\). Finally, take both sides to the power of \(\frac{2}{3}\): \(y = \left(\frac{3}{2}(\sqrt{x}+ C)\right)^{2/3}\). This is the solution for the differential equation.

Key Concepts

Separation of Variables

Separation of variables is a powerful method used to solve differential equations by isolating different types of variables on separate sides of the equation. This is especially useful when dealing with first-order differential equations. The goal is to rearrange the equation such that one variable and its differentials appear on one side, while the other variable and its differentials appear on the other side.

In our example, the given equation is \(2 \sqrt{x y} \frac{d y}{d x} = 1\). Our initial step involves rearranging this into a form where variables \(x\) and \(y\) can be separated. We achieve this by expressing the differential equation in a fraction form \(\frac{d y}{d x} = \frac{1}{2\sqrt{x y}}\). Notice how we successfully separate variables by manipulating the equation to \( \sqrt{y} \cdot dy = \frac{1}{2\sqrt{x}} \cdot dx\). This preparation lays the foundation for integration, leading us to our next method of solving - integration.

Integration

Integration plays a crucial role when solving differential equations after separation of variables. Once the equation is set such that each side exclusively contains either \(x\) or \(y\), we aim to perform integration on both sides.

In our example, we have the equation \(\sqrt{y} \cdot dy = \frac{1}{2\sqrt{x}} \cdot dx\) prepared after separating variables. Now, we proceed to integrate each side:

• Left Side: The integral \(\int \sqrt{y} \, dy\) can be computed as \(\frac{2}{3} y^{3/2}\).
• Right Side: Integrating \(\int \frac{1}{2\sqrt{x}} \, dx\) yields \(\sqrt{x}\).

After integrating, the constants of integration are included as \(C = C_2 - C_1\). Thus, the integrated equation stands at \(\frac{2}{3} y^{3/2} = \sqrt{x} + C\). Through these integrations, we generate expressions that inch closer to the solution.

Solving Differential Equations

Solving a differential equation involves not just separating variables and integrating, but also simplifying the solution to express the dependent variable entirely. Here, we’ve reached the equation \(\frac{2}{3} y^{3/2} = \sqrt{x} + C\) after applying separation of variables and integration.

To solve for \(y\), manipulate both sides by multiplying by \(\frac{3}{2}\), rewriting into \(y^{3/2} = \frac{3}{2} (\sqrt{x} + C)\).
Next, to isolate \(y\), raise both sides to the power of \(\frac{2}{3}\):

• This results in \(y = \left(\frac{3}{2}(\sqrt{x} + C)\right)^{2/3}\).

This expression presents \(y\) purely in terms of \(x\) and arbitrary constants, thus achieving our goal of solving the initial differential equation. This scholarly yet step-by-step methodology is typical for solving differential equations, ensuring that you arrive at a coherent and explanatory solution.