Problem 10 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

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

Step 1: Separate the Variables

The given differential equation is \( \frac{dy}{dx} = x^2 \sqrt{y} \). To solve it, we need to separate the variables. We can do this by dividing both sides by \( \sqrt{y} \) and multiplying both sides by \( dx \), which gives us: \( \frac{1}{\sqrt{y}} \, dy = x^2 \, dx \). Now, the variables are separated.

Step 2: Integrate Both Sides

After separating variables, we integrate both sides. The left side becomes \( \int \frac{1}{\sqrt{y}} \, dy \) and the right side becomes \( \int x^2 \, dx \).- The integral of \( \frac{1}{\sqrt{y}} \) is \( 2\sqrt{y} \).- The integral of \( x^2 \) is \( \frac{x^3}{3} \).Thus, we get: \( 2\sqrt{y} = \frac{x^3}{3} + C \).

Step 3: Solve for y

Now that we have the equation \( 2 \sqrt{y} = \frac{x^3}{3} + C \), we need to solve for \( y \). Divide everything by 2:\[ \sqrt{y} = \frac{x^3}{6} + \frac{C}{2} \].Square both sides to solve for \( y \):\[ y = \left( \frac{x^3}{6} + \frac{C}{2} \right)^2 \].

Key Concepts

Variable Separation

When dealing with differential equations, particularly those that involve multiple variables, a method called variable separation is often employed. This technique is useful when you want to reshape an equation so that each type of variable is isolated on opposite sides.
In our example, we start with the equation \( \frac{dy}{dx} = x^2 \sqrt{y} \). Here, the equation mixes both \( x \) and \( y \) terms. Separating variables requires us to rearrange this into a form that allows integration of each variable independently. We achieved this by dividing both sides by \( \sqrt{y} \) and multiplying by \( dx \), which results in \( \frac{1}{\sqrt{y}} \, dy = x^2 \, dx \).
Now, each variable appears only once on each side: \( y \) on the left and \( x \) on the right, making it possible to integrate with respect to each variable individually. This step is crucial because it simplifies the process of solving complex equations by breaking them into more manageable parts.

Integration

Integration is the mathematical process of finding the integral of a function, and in the context of differential equations, it is essential for solving for a variable. After successfully separating variables, we integrate each side of the equation.
For the given differential equation, the left side becomes \( \int \frac{1}{\sqrt{y}} \, dy \), and the right becomes \( \int x^2 \, dx \).

• The integral of \( \frac{1}{\sqrt{y}} \, dy \) is \( 2\sqrt{y} \). This involves recognizing \( \frac{1}{\sqrt{y}} \) as \( y^{-1/2} \), which integrates to \( 2y^{1/2} \), or \( 2\sqrt{y} \).
• For \( x^2 \, dx \), the integral is \( \frac{x^3}{3} \), using the simple power rule for integration where the general rule is to add one to the exponent and divide by the new exponent.

The result of these integrations gives us the equation \( 2\sqrt{y} = \frac{x^3}{3} + C \), where \( C \) represents the constant of integration. This new equation is now primed for further simplification to find \( y \).

Solving for y

The goal in this step is to express \( y \) entirely in terms of \( x \) and constants, providing us with a solution to the original differential equation.
Starting with the integrated equation \( 2\sqrt{y} = \frac{x^3}{3} + C \), we first isolate \( \sqrt{y} \) by dividing every part of the equation by 2. This gives \( \sqrt{y} = \frac{x^3}{6} + \frac{C}{2} \).
To find \( y \), we need to remove the square root. We do this by squaring both sides, leading to: \[ y = \left( \frac{x^3}{6} + \frac{C}{2} \right)^2 \].
In this expression, \( y \) is now isolated, demonstrating a direct relationship to \( x \), fully solving the differential equation. This final step ties together the entire process of separating, integrating, and solving for the variable we are interested in, providing a complete understanding of the original problem.