Problem 49 - Calculus and its applications

Question

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is inversely proportional to the cube of \(y\).

Step-by-Step Solution

Answer

The differential equation is \( \frac{dy}{dx} = \frac{1}{y^3} \). The general solution is \( y = (4x + C')^{1/4} \).

Step 1: Identify the relationship

The problem states that the rate of change of \( y \) with respect to \( x \) is inversely proportional to the cube of \( y \). This can be expressed as \( \frac{dy}{dx} = \frac{k}{y^3} \). According to the problem, we set \( k = 1 \).

Step 2: Write the differential equation

Using the proportionality, the differential equation becomes \( \frac{dy}{dx} = \frac{1}{y^3} \).

Step 3: Solve the differential equation

To solve \( \frac{dy}{dx} = \frac{1}{y^3} \), separate the variables: \( y^3 \, dy = dx \). Integrate both sides: \[ \int y^3 \, dy = \int dx \] This yields \( \frac{y^4}{4} = x + C \), where \( C \) is the integration constant.

Step 4: Express the general solution

Rearrange the expression \( \frac{y^4}{4} = x + C \) to find \( y \) in terms of \( x \). Multiply through by 4: \[ y^4 = 4x + 4C \] Thus, the general solution is \( y = (4x + C')^{1/4} \), where \( C' = 4C \).

Step 5: Apply initial condition (if given)

If an initial condition is provided, such as \( y(x_0) = y_0 \), substitute these values into the general solution to solve for \( C' \). This gives the particular solution.

Key Concepts

Proportional Relationships

In mathematics, proportional relationships describe a situation where two quantities maintain a constant ratio. These relationships can be either direct or inverse.
- **Direct proportionality**: Here, one quantity increases as the other increases. It is expressed as \( y = kx \), where \( y \) and \( x \) are the quantities, and \( k \) is the constant of proportionality.
- **Inverse proportionality**: In this case, one quantity increases as the other decreases, and it is represented by \( y = \frac{k}{x} \).
In the given exercise, it is important to recognize that the rate of change of \( y \) with respect to \( x \) is inversely proportional to the cube of \( y \). Therefore, you should express this relationship as \( \frac{dy}{dx} = \frac{k}{y^3} \). Since we assume \( k = 1 \), the differential equation becomes \( \frac{dy}{dx} = \frac{1}{y^3} \). This sets the stage for solving the differential equation using further calculus techniques.

Separation of Variables

Separation of variables is a method used to solve differential equations. It involves rearranging the differential equation to isolate each variable on opposite sides of the equation.
In the exercise, once you've established that \( \frac{dy}{dx} = \frac{1}{y^3} \), separation of variables allows you to move all terms involving \( y \) to one side and terms involving \( x \) to the other side, resulting in \( y^3 \, dy = dx \).
This form makes it easier to integrate, as you can now consider each side of the equation independently. The concept here is to treat \( dy \) and \( dx \) as separate entities, solving for \( y \) and \( x \) by integrating both sides.
The separation of variables technique is particularly useful in this exercise because it transforms a complex differential equation into a more manageable form. Successfully using this method helps in finding the general and potentially particular solutions of the equation.

Integral Calculus

Integral calculus is a branch of calculus focused on the accumulation of quantities and the area under curves. In the context of differential equations, it is used to find antiderivatives, which are necessary to solve equations involving derivatives.
After separating variables in our exercise, the next step is to integrate both sides. We perform the operations: \[\int y^3 \, dy = \int dx\]
This integration results in \( \frac{y^4}{4} = x + C \), where \( C \) is the constant of integration. Here, integral calculus allows us to transition from the derivative \( \frac{dy}{dx} \) to a relationship between \( y \) and \( x \).
Rewriting this equation by solving for \( y \), you find \( y = (4x + C')^{1/4} \), where \( C' \) is an adjustment of the integration constant. This manipulation provides the general solution to our original differential equation.
Integral calculus is therefore essential for understanding and solving differential equations by providing the methodologies necessary to integrate and find solutions. This technique, combined with our prior steps, brings us closer to a complete understanding of the problem at hand.