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Magazine Chapter 19: Problem 10
Solve the differential equation. \(y^{\prime}+y \tan x=\sin x\)
Short Answer
Expert verified
The solution is \(y = \cos x (\ln |\sec x + \tan x| + C)\).
Step by step solution
01
Write the Given Differential Equation
The given differential equation is \( y' + y \tan x = \sin x \). This is a first-order linear differential equation.
02
Identify the Standard Form
The standard form of a first-order linear differential equation is \( y' + P(x)y = Q(x) \). Here, \( P(x) = \tan x \) and \( Q(x) = \sin x \).
03
Find the Integrating Factor
The integrating factor, \( \, \mu(x) \, \), is given by \( e^{\int P(x) \, dx} \). So, we calculate:\[\mu(x) = e^{\int \tan x \, dx} = e^{\ln |\sec x|} = \sec x\]
04
Multiply Through by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( \sec x \):\[\sec x \, y' + y \sec x \tan x = \sin x \sec x\]This simplifies to:\[\frac{d}{dx}(y \sec x) = \sin x \sec x\]
05
Integrate Both Sides
Integrate both sides:\[\int \frac{d}{dx}(y \sec x) \, dx = \int \sin x \sec x \, dx\]\(y \sec x = \ln |\sec x + \tan x| + C\)
06
Solve for y
Solve for \(y\) by dividing by \(\sec x\):\[y = \cos x (\ln |\sec x + \tan x| + C)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation involves derivatives of the first degree without any higher power terms or products. These types of equations have a certain form which makes them easier to handle compared to non-linear equations. If the equation can be written as
This form allows us to explore useful techniques such as finding an integrating factor. Knowing what qualifies as a first-order linear differential equation is the first step to successfully finding a solution. Recognizing this can save you time and unnecessary calculations, ensuring you apply the correct solution methods.
- \( y' + P(x)y = Q(x) \)
This form allows us to explore useful techniques such as finding an integrating factor. Knowing what qualifies as a first-order linear differential equation is the first step to successfully finding a solution. Recognizing this can save you time and unnecessary calculations, ensuring you apply the correct solution methods.
Integrating Factor Method
The integrating factor method is a smart approach used to solve first-order linear differential equations. It involves multiplying the entire differential equation by a special function called the 'integrating factor' to simplify the equation. The integrating factor is given by
In our exercise, \( P(x) \) was \( \tan x \). After performing the integral of \( \tan x \), the integrating factor turned out to be \( \sec x \).
- \( \mu(x) = e^{\int P(x) \ dx} \)
In our exercise, \( P(x) \) was \( \tan x \). After performing the integral of \( \tan x \), the integrating factor turned out to be \( \sec x \).
- Finding the correct integrating factor is crucial as it simplifies the differential equation into a form that can be easily integrated.
- The main goal is to rewrite the equation so that the derivative becomes evident, allowing straightforward integration.
Calculus Problem Solving
Calculus serves as a mighty tool to tackle differential equations by applying derivative and integration principles. The heart of calculus problem-solving lies in understanding
We used the concept of integration to solve the transformed equation
For effective calculus problem-solving:
- how to manipulate expressions using derivatives,
- how integrals provide solutions to these problems.
We used the concept of integration to solve the transformed equation
- \( \frac{d}{dx}(y \sec x) = \sin x \sec x \)
For effective calculus problem-solving:
- Always simplify expressions when possible.
- Recognize patterns that match known integral formulas.
- Perform algebraic manipulations to clean up complex expressions.
Solving Differential Equations
Solving differential equations involves multiple steps, but understanding the basic foundation helps a lot. The fundamental goal is to find a function that satisfies the given equation. Here's the general approach we used in the given example:
**Identify the Type of Differential Equation**
Effective practice and applying these steps ensures that you become proficient at solving similar problems quickly and accurately.
**Identify the Type of Differential Equation**
- Determine whether it is linear or non-linear.
- Write it in the standard form if possible.
- This allows for simplifying the equation.
- Use integration to find the specific integrating factor.
- Transform the equation to a recognizable integration form.
- Perform integrations on both sides.
Effective practice and applying these steps ensures that you become proficient at solving similar problems quickly and accurately.
