Chapter 2: Problem 45
In Problems \(41-50\) solve the given differential equation subject to the indicated initial condition. $$ y^{\prime}+(\tan x) y=\cos ^{2} x, \quad y(0)=-1 $$
Short Answer
Expert verified
The solution is \( y = \cos x (\sin x - 1) \) with initial condition \( y(0) = -1 \).
Step by step solution
01
Identify the Type of Equation
The given differential equation is a first-order linear differential equation of the form \( y' + P(x)y = Q(x) \). Here, \( P(x) = \tan x \) and \( Q(x) = \cos^2 x \). The initial condition provided is \( y(0) = -1 \).
02
Find the Integrating Factor
To solve the first-order linear differential equation, we need to find the integrating factor \( \mu(x) \). The integrating factor is found from \( \mu(x) = e^{\int P(x) \; dx} = e^{\int \tan x \; dx} \).The integral of \( \tan x \) is \( \ln |\sec x| \), so the integrating factor is \( \mu(x) = |\sec x| \).
03
Simplify the Integrating Factor
Since the absolute value is not an issue for the interval over which the solution applies, we use \( \mu(x) = \sec x \). This will simplify the equation without changing the solution.
04
Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by \( \sec x \), resulting in:\[\sec x \cdot y' + \sec x \cdot \tan x \cdot y = \cos^2 x \cdot \sec x\]This simplifies to \( y' \cdot \sec x + y \cdot \sec x \cdot \tan x = \cos x \), which can be rewritten for further simplification.
05
Recognize the Left Side as a Derivative
The left-hand side can be rewritten as the derivative of \( y \cdot \sec x \):\[\frac{d}{dx}(y \cdot \sec x) = \cos x\]
06
Integrate Both Sides
Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(y \cdot \sec x) \, dx = \int \cos x \, dx\]This yields:\[y \cdot \sec x = \sin x + C\]
07
Solve for y
Solve for \( y \) by isolating it:\[y = \sin x \cdot \cos x + C \cdot \cos x\]
08
Apply the Initial Condition
Use the initial condition \( y(0) = -1 \) to solve for \( C \).Substitute \( x = 0 \) and \( y = -1 \) into the equation:\[-1 = \sin(0) \cdot \cos(0) + C \cdot \cos(0) \]This simplifies to \(-1 = 0 + C \cdot 1\), leading to \( C = -1 \).
09
Write the General Solution
Substitute \( C = -1 \) back into the formula for \( y \):\[y = \sin x \cdot \cos x - \cos x\]This simplifies to:\[y = \cos x (\sin x - 1)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating factor
In first-order linear differential equations, the integrating factor is a special function used to make the equation easy to solve. It is particularly useful when your equation looks like this:
Since the absolute value does not affect the interval of the solution, \( \mu(x) = \sec x \) becomes our integrating factor. This crucial step transforms the differential equation into a solvable form.
- \( y' + P(x)y = Q(x) \)
- \( \mu(x) = e^{\int P(x) \, dx} \)
Since the absolute value does not affect the interval of the solution, \( \mu(x) = \sec x \) becomes our integrating factor. This crucial step transforms the differential equation into a solvable form.
Initial condition
An initial condition is a requirement that specifies the value of the solution at a particular point. It ensures the solution to the differential equation is unique. For example, in our problem, the initial condition given is:
In this exercise, after finding the general solution of the original differential equation, we substitute \( x = 0 \) and \( y = -1 \) back into it. This allows us to solve for the constant \( C \), ensuring our solution satisfies the given initial condition.
- \( y(0) = -1 \)
In this exercise, after finding the general solution of the original differential equation, we substitute \( x = 0 \) and \( y = -1 \) back into it. This allows us to solve for the constant \( C \), ensuring our solution satisfies the given initial condition.
Solution method
Solving a first-order linear differential equation typically involves several key steps. Here's an overview of the method used in this example:
- Identify the structure: Ensure the equation is in the form \( y' + P(x)y = Q(x) \).
- Calculate the integrating factor \( \mu(x) \): This turns the equation into a product rule form.
- Multiply every term by the integrating factor to transform the equation.
- Recognize the transformed left side as a derivative of a product \( \frac{d}{dx}(y \cdot \mu(x)) \).
- Integrate both sides of this equation for a complete solution.
- Apply the initial condition to find any unknown constants.
Differential equation simplification
Simplifying a differential equation is about reformulating it into a simpler form that reveals its solution more readily. The original equation:
This systematic simplification takes seemingly complicated equations and turns them into more manageable expressions, making the solving process much more straightforward.
- \( y' + \tan x \cdot y = \cos^2 x \)
- \( \sec x \cdot y' + (-\sec x \cdot \tan x) \cdot y = \cos x \)
This systematic simplification takes seemingly complicated equations and turns them into more manageable expressions, making the solving process much more straightforward.