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Chapter 11: Problem 2

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d P}{d t}=-2 P, \quad P(0)=1$$

Short Answer

Expert verified
The solution is \( P(t) = e^{-2t} \).

Step by step solution

01

Write the differential equation

We start with the given differential equation \( \frac{dP}{dt} = -2P \). This is a first-order linear ordinary differential equation.
02

Separate the variables

To separate the variables, divide both sides of the equation by \(P\) and multiply both sides by \(dt\): \( \frac{dP}{P} = -2 \, dt \). Now the variables are separated, with all terms involving \(P\) on one side and terms involving \(t\) on the other side.
03

Integrate both sides

Integrate both sides of the equation: \( \int \frac{dP}{P} = \int -2 \, dt \). The left side integrates to \(\ln |P|\) and the right side integrates to \(-2t + C\), where \(C\) is the integration constant. We have \( \ln |P| = -2t + C \).
04

Solve for P

To solve for \(P\), we exponentiate both sides to remove the logarithm: \( e^{\ln |P|} = e^{-2t + C} \). This simplifies to \( |P| = e^{C} e^{-2t} \). Let \( C_1 = e^{C} \) (a positive constant), then \( P = C_1 e^{-2t} \).
05

Apply initial conditions

Use the initial condition \( P(0) = 1 \). Substitute \(t = 0\) and \(P = 1\) into the equation: \( 1 = C_1 e^{0} \). Thus, \( C_1 = 1 \). This gives us the specific solution: \( P(t) = e^{-2t} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
Separation of variables is a method used to solve ordinary differential equations (ODEs). It works by rearranging an equation so that each variable and its derivative are on opposite sides of the equation.

This allows the integration of each side individually. In the given exercise, we start with the differential equation \( \frac{dP}{dt} = -2P \). For separation of variables, we manipulate this into \( \frac{dP}{P} = -2 \, dt \), which has isolated all \(P\) variables on one side and \(t\) variables on the other.

When this setup is achieved, we can proceed to solve the equation by integrating each side separately. This leads us into the next step in problem-solving.
Initial Conditions
Initial conditions help to identify a specific solution from a family of possible solutions given by a differential equation.

Every time we integrate, we add a constant \(C\). The solution includes a constant because the differentiation of a constant is zero, making it seem invisible in the original differential equation.

In the provided exercise, the initial condition is \(P(0) = 1\). This condition tells us the value of \(P\) when \(t\) is zero.
  • We use it to determine the numerical value of the constant \(C_1\).
  • After calculating, \(C_1 = 1\), we integrate it into our final solution.
Without applying initial conditions, the solution might remain too general, and thus, less useful for real-world problems.
Ordinary Differential Equations
Ordinary differential equations (ODEs) involve functions of a single independent variable and their derivatives. These types of equations are widely used in various fields such as physics, engineering, and biology to describe continuous processes.

The provided exercise deals with the ODE \( \frac{dP}{dt} = -2P \), which is a first-order ODE. This means it involves the first derivative of \(P\) with respect to \(t\).

First-order ODEs are simpler than higher-order types but remain highly significant in modeling exponential growth or decay scenarios, such as population dynamics or radioactive decay.
Integration
Integration is the process of finding a function given its derivative, often seen as the opposite of differentiation. In solving differential equations through separation of variables, integration pulls us out of the derivative-dominated world.

In the exercise, after separating the variables, we integrate both sides:
  • The left-hand side, \( \int \frac{dP}{P} \), equates to \( \ln |P| \) as per the basic integration rules.
  • The right-hand side, \( \int -2 \, dt \), solves to \(-2t + C\).
These integrations lead us to the form \( \ln |P| = -2t + C \).
Integration brings about a few defining constants (in this case \(C\)), which are crucial in obtaining the full solution in conjunction with initial conditions.

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Most popular questions from this chapter

Give an example of: A differential equation whose slope field has all the slopes positive.

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. The half-life of hydrocodone bitartrate in the body is 3.8 hours, and the usual oral dose is 10 mg. (a) Write a differential equation for the quantity, \(Q,\) of hydrocodone bitartrate in the body at time \(t,\) in hours since the drug was fully absorbed. (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve the differential equation given in part (a). (d) Use the half-life to find the constant of proportionality, \(k\) (e) How much of the 10 mg dose is still in the body after 12 hours?

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