Problem 69 - Calculus Single Variable

Question

The pressure \(P\) and temperature \(T\) in the outer envelope of a white dwarf (star) are related by the differential equation $$ \frac{d P}{d T}=C \frac{T^{7.5}}{P}, P(0)=0 $$ where \(C\) is a constant. Find \(P\) as a function of \(T\).

Step-by-Step Solution

Answer

The pressure as a function of temperature is \( P(T) = \sqrt{\frac{2C}{8.5}} T^{4.25} \).

Step 1: Separate the Variables

Start by rearranging the given differential equation \( \frac{d P}{d T}=C \frac{T^{7.5}}{P} \) to isolate variables involving \( P \) on one side and those involving \( T \) on the other side. Multiply both sides by \( P \) and \( dT \) to obtain \( P \, dP = C \, T^{7.5} \, dT \).

Step 2: Integrate Both Sides

Integrate both sides of the equation. The integration on the left-hand side with respect to \( P \) is \( \int P \, dP = \frac{P^2}{2} \) and on the right-hand side with respect to \( T \), it is \( \int C \, T^{7.5} \, dT = C \frac{T^{8.5}}{8.5} + D \), where \(D\) is the constant of integration.

Step 3: Solve for Constant of Integration

Since the initial condition given is \( P(0) = 0 \), substitute this into the integrated equation: \( \frac{0^2}{2} = C \frac{0^{8.5}}{8.5} + D \). This simplifies to \( D = 0 \).

Step 4: Solve for P

Now, write the equation obtained from the integration with known constant \(D\): \( \frac{P^2}{2} = C \frac{T^{8.5}}{8.5} \). Solve for \(P\): \( P = \sqrt{\frac{2C}{8.5} T^{8.5}} = \sqrt{\frac{2CT^{8.5}}{8.5}} \). This gives the relationship \( P(T) = \sqrt{\frac{2C}{8.5}} T^{4.25} \).

Key Concepts

Separation of Variables

Separation of variables is a fundamental method used to solve certain kinds of differential equations. When an equation involves two variables, say \( P \) and \( T \), you can often separate the variables by manipulating the equation into two distinct parts, each containing only one of the variables.
This technique is particularly useful when dealing with equations that can be rearranged so that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side.
In our exercise, we separated the variables by multiplying both sides of the equation by \( P \) and \( dT \), which rearranged the equation into:

• \( P \, dP = C \, T^{7.5} \, dT \)

This separation allows us to solve the differential equation by integrating each side individually. By isolating the variables, it becomes easier to perform further analytical operations such as integration.

Integration

Integration is the process of finding the integral of a function, which is essentially the opposite of differentiating. It is a vital step in solving differential equations after variables have been separated. There are two main components when integrating: the integral function and the constant of integration.
In the context of the exercise, integrating both sides of the equation was the next logical step after separating the variables. Each side of the equation correlates to a different variable, and we integrate separately with respect to each variable.
For example, the integration process on the left side for \( P \) includes the integral:

• \( \int P \, dP = \frac{P^2}{2} \)

On the right side for \( T \), we have:

• \( \int C \, T^{7.5} \, dT = C \frac{T^{8.5}}{8.5} + D \)

Here, \( D \) represents the constant of integration, which is determined using initial conditions.

Initial Value Problem

An initial value problem is a type of differential equation that provides additional conditions, known as initial conditions, to determine a unique solution. These conditions specify the value of the unknown function at a specific point, which helps in finding the exact form of the solution.
In the given exercise, the initial condition provided is \( P(0) = 0 \). This means that when the temperature \( T \) is zero, the pressure \( P \) should also be zero.
The initial condition is crucial for determining the constant of integration. By substituting \( T = 0 \) into our integrated function, we find:

• \( \frac{0^2}{2} = C \frac{0^{8.5}}{8.5} + D \)

This simplifies to \( D = 0 \), effectively removing \( D \) from the equation, and allowing us to solve for \( P \). Precise initial values ensure the solution caters to specific scenarios, making the answer both practical and relevant.