Problem 38

Question

Separating variables, we obtain $d P / P=k \cos t d t,$ so $$\ln |P|=k \sin t+c \quad \text { and } \quad P=c_{1} e^{k \sin t}$$ If $P(0)=P_{0},$ then $c_{1}=P_{0}$ and $P=P_{0} e^{k \sin t}$.

Step-by-Step Solution

Verified

Answer

$ P = P_0 e^{k \sin t} $ with $ c_1 = P_0 $.

1Step 1: Understanding the Differential Equation

We start with the differential equation $ \frac{dP}{P} = k \cos t \ dt $. Our task is to find the general solution for $ P $.

2Step 2: Separate the Variables

The given equation can be rearranged to isolate the differentials by multiplying both sides by $ P $ and dividing by $ \cos t $, resulting in $ \frac{dP}{P} = k \cos t \ dt $.

3Step 3: Integrate Both Sides

Integrate the left-hand side with respect to $ P $ and the right-hand side with respect to $ t $:\[ \int \frac{1}{P} \, dP = \int k \cos t \, dt \] which results in $ \ln |P| = k \sin t + C $.

4Step 4: Solve for P

To find $ P $, we exponentiate both sides of the equation to solve for $ P $:\[ P = e^{k \sin t + C} = e^C e^{k \sin t} \] Let $ c_1 = e^C $, thus $ P = c_1 e^{k \sin t} $.

5Step 5: Apply Initial Condition

Given the initial condition $ P(0) = P_0 $, substitute $ t = 0 $ and $ P = P_0 $ into the equation:\[ P_0 = c_1 e^{k \sin 0} = c_1 \times 1 = c_1 \] Hence, $ c_1 = P_0 $.

6Step 6: Final Solution Expression

Substitute $ c_1 = P_0 $ back into the expression for $ P $:\[ P = P_0 e^{k \sin t} \] This is the expression for $ P $ as a function of $ t $.

Key Concepts

Separation of VariablesInitial ConditionExponential FunctionIntegration Techniques

Separation of Variables

Separation of Variables is a fundamental method used when solving differential equations, especially useful for equations that can be restructured to isolate functions of a variable on each side. This technique allows you to take a differential equation and rewrite it so that each side only contains one of the variables multiplied by its differential. For example, in the equation $ \frac{dP}{P} = k \cos t \, dt $, the goal of separation is to get all terms involving $ P $ on one side and those involving $ t $ on the other. This simplifies the solution process significantly.The process involves a few basic steps:

Identify the terms that can be separated.
Rearrange the equation, moving all terms involving the same variable to one side.
Integrate each side independently. Once the variables are separated, it becomes possible to integrate each side, helping us to find a function solution.

Initial Condition

An initial condition in differential equations specifies particular values for the solution at a given point, often used to find the specific form of the solution. It helps in determining the constant of integration, $ c_1 $, when you solve a differential equation generically. For instance, in our example, we apply the initial condition $ P(0) = P_0 $. This means at $ t = 0 $, the value of $ P $ is known. Substitute $ t = 0 $ and $ P = P_0 $ into your solution for $ P $ to find $ c_1 $, which lets us express the solution uniquely.Initial conditions are crucial because, without them, a differential equation could have infinitely many solutions. They help tailor the solution to fit specific scenarios, providing particularity and relevance to the real-world phenomena being modeled.

Exponential Function

The exponential function occurs frequently in solutions to differential equations, especially in the form $ e^{kx} $. It represents processes that change proportionally at a constant rate, such as growth or decay, modeled by this influential curve.In the context of our exercise, $ P = c_1 e^{k \sin t} $ shows $ P $ growing or decaying with respect to $ \sin t $. Here, $ k $ modulates the rate of exponential change, while $ \sin t $ introduces oscillation into the dynamic.The exponential function has key properties:

Its derivative is proportional to the function itself. This makes it particularly adaptive to modeling naturally occurring exponential behavior.
It always stays positive, making it a powerful tool for scenarios like population growth where negative values aren't practical.

Integration Techniques

Integration techniques are essential for solving the separated sides of a differential equation. Once variables are separated, as seen in earlier sections, integration comes next to find the function itself.One technique is integrating standard forms. For instance, the integral of $ \frac{1}{P} \, dP $ typically transforms into the natural logarithm, $ \ln |P| $. Similarly, $ \int \cos t \, dt $ straightforwardly becomes $ \sin t + C $ with respect to $ t $, where $ C $ is the constant of integration.Learning integration requires familiarity with various techniques:

Substitution, where a variable is changed to simplify the integration.
Integration by parts when tackling products of different functions.
Recognizing integrable forms instantly can save time and simplify processes.

Integrating effectively is a skillful art, crucial in mastering the solutions of many problems in calculus, especially differential equations.

Problem 37

Problem 38

Other exercises in this chapter

Problem 37

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Problem 37

We note that $\left(M_{y}-N_{x}\right) / N=2 x /\left(4+x^{2}\right),$ so an integrating factor is \(e^{-2 \int x d x /\left(4+x^{2}\right)}=1 /\left(4+x^{2}\

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Problem 38

Write the differential equation as $d P / d t-a P=-b P^{2}$ and let $u=P^{-1}$ or $P=u^{-1}$. Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substi

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Problem 38

(a) The second derivative of $y$ is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution c

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