Problem 57 - Calculus Early Transcendentals

Question

Solve the following problems using the method of your choice. $$\frac{d p}{d t}=\frac{p+1}{t^{2}}, p(1)=3$$

Step-by-Step Solution

Answer

Question: Solve the first-order ordinary differential equation (ODE) given by \(\frac{d p}{d t}=\frac{p+1}{t^{2}}\) with the initial condition \(p(1)=3\). Solution: The solution of the given ODE is \(p(t) = e^{-\frac{1}{t} + \ln{4}} - 1\).

Step 1: Identify the method to solve the ODE

Separation of Variables is a useful method to solve first-order ODEs. This method involves separating the variables (in this case, \(p\) and \(t\)) on each side of the equation and then integrating both sides.

Step 2: Separate the variables

To separate the variables, we need to rewrite the given ODE in the form \(\frac{d p}{p + 1} = \frac{d t}{t^2}\). We can do this by multiplying both sides by \(t^2\) and dividing both sides by \((p + 1)\): $$\frac{d p}{p + 1} = \frac{d t}{t^2}.$$

Step 3: Integrate both sides of the equation

Now, we must integrate both sides of the equation with respect to their respective variables: $$\int \frac{d p}{p +1} = \int \frac{d t}{t^2}.$$ Let's integrate each side separately. For the left side, since the integral of \(\frac{1}{p+1}\) is \(\ln|p+1|\), we have: $$\int \frac{d p}{p +1} = \ln|p+1|+C_1.$$ For the right side, the integral of \(\frac{1}{t^2}\) is \(-\frac{1}{t}\), thus: $$\int \frac{d t}{t^2} = -\frac{1}{t} + C_2.$$ Putting both results together, we have: $$\ln|p+1|+C_1 = -\frac{1}{t} + C_2.$$

Step 4: Solve for \(p\)

Now, we must isolate \(p\) on one side of the equation. First, subtract \(C_1\) from both sides: $$\ln|p+1| = -\frac{1}{t} + C_2 - C_1.$$ Let \(C = C_2 - C_1\), so our equation becomes: $$\ln|p+1| = -\frac{1}{t} + C.$$ Now, exponentiate both sides to isolate \(p+1\): $$p+1 = e^{-\frac{1}{t} + C}.$$ Finally, simplify for \(p\): $$p(t) = e^{-\frac{1}{t} + C}-1.$$

Step 5: Use the initial condition to find the constant \(C\)

We are given the initial condition \(p(1) = 3\). Plugging this into our equation for \(p(t)\), we can solve for \(C\): $$3 = e^{-\frac{1}{1} + C} - 1.$$ Adding \(1\) to both sides, we obtain: $$4 = e^{C}.$$ Taking the natural logarithm of both sides, we find the constant \(C\): $$C = \ln{4}.$$

Step 6: Write the final solution

Now that we have found the constant \(C\), we can write the final solution for the ODE: $$p(t) = e^{-\frac{1}{t} + \ln{4}} - 1.$$

Key Concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) play a critical role in mathematics when it comes to modeling how things change over time or space. ODEs are equations that involve functions and their derivatives. Here, an equation is deemed "ordinary" if it consists of da single independent variable.
One of the most fundamental forms of ODEs is the first-order ODE, which involves first derivatives. For example, in the given exercise, the equation \( \frac{d p}{d t}=\frac{p+1}{t^{2}} \) is a first-order ODE since it involves the first derivative of \( p \) with respect to \( t \).
When solving these equations, our goal is to find the unknown function, which in this case is \( p(t) \), that satisfies the equation.

• Before solving an ODE, it's helpful to identify any given conditions, such as initial conditions, which help pinpoint the exact solution among many potential ones.
• The solution of an ODE can often be visualized as a curve or a family of curves on a graph.
• Each solution's curve represents the trajectory of the system described by the ODE across time or space.

Today, we frequently solve ODEs using numerical methods or algebraic manipulations, such as separation of variables, to derive an analytical solution.

Integration

Integration is the mathematical process of finding the antiderivative of a function. In the context of solving ODEs, integration is often used to find a function that describes an infinite collection of curves that meet specific criteria.
When we talk about separation of variables, it's a technique used for solving ODEs by rearranging an equation so that each variable occurs exclusively on one side of the equation. This makes it straightforward to integrate each side independently.

• In our example, \( \frac{d p}{p + 1} = \frac{d t}{t^2} \) is separated so that \( p \,\text{and}\, t \) only appear with their respective differentials.
• Integrating \( \int \frac{d p}{p+1} \) gives \( \ln |p+1| \), and \( \int \frac{d t}{t^2} \) results in \( -\frac{1}{t} \).
• Proper integration allows us to determine a general solution to the ODE, which includes an arbitrary constant \( C \), representing the infinite solutions that exist without a specific initial condition.

Each integration step helps peel back layers of complexity to simplify the equation down to the unknown function's characteristics.

Initial Condition

An initial condition is a crucial piece of information that allows us to determine the specific solution to an ODE that fits a real-world scenario. Without it, we get a family of functions due to the arbitrary constant arising in our integration.
Initial conditions specify the value of the unknown function at a certain point and help eliminate excess possible solutions. In the given problem, the initial condition is \( p(1) = 3 \). This acts like a point of reference.

• Apply the initial condition after integrating and simplifying the general solution. Doing this allows you to solve for the constant \( C \).
• In our example, substituting \( p(1)=3 \) into the equation \( 3 = e^{-\frac{1}{1} + C} - 1 \) helps us calculate \( C \) as \( \ln 4 \).
• After finding \( C \), rewrite the general solution using this constant to get the explicit solution for the differential equation.

This way, we transition from a general framework that applies to numerous possibilities to a specific path perfectly tailored to the problem at hand.