Problem 10 - Thomas Calculus

Question

Solve the differential equations in Exercises \(9-22\) $$\frac{d y}{d x}=x^{2} \sqrt{y}, \quad y>0$$

Step-by-Step Solution

Answer

The solution is \( y = \left( \frac{x^3}{6} + C \right)^2 \), where \( C \) is a constant.

Step 1: Separate Variables

Start by separating the variables, so all terms involving \( y \) are on one side, and all terms involving \( x \) are on the other side. Rewrite the equation as: \[ \frac{dy}{\sqrt{y}} = x^2 \, dx \]

Step 2: Integrate Both Sides

Integrate both sides of the equation. On the left side, integrate with respect to \( y \), and on the right side, integrate with respect to \( x \). The integrals become: \[ \int \frac{dy}{\sqrt{y}} = \int x^2 \, dx \] which simplifies to \[ 2\sqrt{y} = \frac{x^3}{3} + C \] where \( C \) is the constant of integration.

Step 3: Solve for \( y \)

Square both sides of the equation to solve for \( y \). This gives: \[ (\sqrt{y})^2 = \left( \frac{x^3}{6} + C^2 \right)^2 \] thus simplifying to \( y = \left( \frac{x^3}{6} + C \right)^2 \).

Key Concepts

Variable Separation

The concept of variable separation is a fundamental technique used in solving differential equations. The goal is to rearrange the equation so that each variable appears on its own side. This step is crucial because it allows for the equation to be integrated with respect to each independent variable.In our example, we start with the equation \(\frac{dy}{dx} = x^2 \sqrt{y}\). To separate the variables, we move all terms involving \(y\) to one side of the equation and all terms involving \(x\) to the other. We achieve this by rewriting the equation as \(\frac{dy}{\sqrt{y}} = x^2 \, dx\).

• Commandments for successful separation: Isolate variables.
• Ensure both sides contain only one variable.

Once variables are separated, the road to integration becomes clear. This method is a powerful tool because it reduces the complexity of solving the equation.

Integration

Integration is the next step after successfully separating the variables. It is essentially the process of finding the antiderivative of a function. In the context of differential equations, integration will help us find the general solution.For our separated equation \(\frac{dy}{\sqrt{y}} = x^2 \ dx\), integrating both sides gives:

• Left Side: \(\int \frac{dy}{\sqrt{y}}\) becomes \(2\sqrt{y}\).
• Right Side: \(\int x^2 \, dx\) results in \(\frac{x^3}{3}\).

Both sides have now been integrated, resulting in the equation \(2\sqrt{y} = \frac{x^3}{3} + C\). The integration process brings the equation closer to a form where we can solve for \(y\). Each integration increases our understanding of the relationship between \(x\) and \(y\).

Constant of Integration

The constant of integration, often denoted as \(C\), appears as a result of the integration process. It represents an unknown constant that can take any real value. This constant is crucial because when integrating, you're essentially solving for a family of functions that all work as solutions to the differential equation.In our example, after integrating both sides, we get the equation \(2\sqrt{y} = \frac{x^3}{3} + C\). The \(C\) can be any constant, which means there are potentially infinite solutions fitting various conditions:

• The constant of integration gives us flexibility.
• It helps find particular solutions when extra conditions are given.

Understanding the role of this constant is essential for fully solving and applying differential equations in real-world scenarios.