Chapter 1: Problem 7
Solve \(y^{\prime}+\cos (x) y=\cos (x)\).
Short Answer
Expert verified
The solution is \( y(x) = 1 + Ce^{-\sin(x)} \).
Step by step solution
01
Identify the Type of Differential Equation
The given equation is \( y' + \cos(x) y = \cos(x) \). This is a first-order linear differential equation of the form \( y' + P(x)y = Q(x) \) where \( P(x) = \cos(x) \) and \( Q(x) = \cos(x) \).
02
Compute the Integrating Factor
The integrating factor \( \mu(x) \) is given by the formula \( e^{\int P(x) \, dx} \). Here, \( P(x) = \cos(x) \), so we need to compute the integral:\[\int \cos(x) \, dx = \sin(x)\]Thus, the integrating factor is \( \mu(x) = e^{\sin(x)} \).
03
Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( e^{\sin(x)} \):\[e^{\sin(x)}(y' + \cos(x)y) = e^{\sin(x)} \cos(x)\]This simplifies to:\[(y \cdot e^{\sin(x)})' = \cos(x) \cdot e^{\sin(x)}\]
04
Integrate Both Sides
Now integrate both sides with respect to \( x \):\[\int (y \cdot e^{\sin(x)})' \, dx = \int \cos(x) \cdot e^{\sin(x)} \, dx\]The left-hand side integrates to \( y \cdot e^{\sin(x)} \). Use substitution for the right-hand side: let \( u = \sin(x) \), \( du = \cos(x) \, dx \), so:\[\int e^{u} \, du = e^{u} + C\]Substituting back, we get \( e^{\sin(x)} + C \).
05
Solve for y
From the integration, we have:\[y \cdot e^{\sin(x)} = e^{\sin(x)} + C\]Now divide both sides by \( e^{\sin(x)} \) to solve for \( y \):\[y = 1 + Ce^{-\sin(x)}\]
06
State the Solution
The general solution to the differential equation is:\[ y(x) = 1 + Ce^{-\sin(x)} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating factor
In solving first-order linear differential equations, the integrating factor method becomes an essential tool. Here, the purpose of the integrating factor is to transform a differential equation into an exact equation. This means we convert the original equation into one where the left-hand side can be expressed as the derivative of a product.
The integrating factor \( \mu(x) \) for a differential equation of the form \( y' + P(x)y = Q(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). This factor simplifies the process of finding the solution.
The integrating factor \( \mu(x) \) for a differential equation of the form \( y' + P(x)y = Q(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). This factor simplifies the process of finding the solution.
- To calculate the integrating factor, you first identify \( P(x) \).
- Next, compute the integral of \( P(x) \) to find the exponent for the integrating factor.
- In our example, \( P(x) = \cos(x) \), so the integrating factor is \( e^{\sin(x)} \).
General solution
Finding the general solution involves transforming the differential equation to help isolate \( y \). Once an equation is multiplied by the integrating factor, it can often be rewritten as a simple derivative of a product.
This step is crucial because it allows us to integrate directly.
This step is crucial because it allows us to integrate directly.
- For the given problem, after multiplying by the integrating factor, the equation becomes \((y \cdot e^{\sin(x)})' = \cos(x) \cdot e^{\sin(x)}\).
- The left side represents the derivative of \( y \cdot e^{\sin(x)} \), allowing for simple integration.
- When integrated over \( x \), both sides yield functions plus an arbitrary constant \( C \).
Substitution method
The substitution method in solving differential equations is a supportive strategy that is especially useful in handling integrals complicating our work. In our context, substitution eases the integration of the right-hand side of the simplified equation.
The key steps involve:
The key steps involve:
- Selecting a substitution that simplifies the integral. Here, \( u = \sin(x) \).
- Transforming the differential \( dx \) into \( du \) using the derivative \( du = \cos(x) \, dx \).
- The troublesome integral \( \int \cos(x) \cdot e^{\sin(x)} \, dx \) becomes \( \int e^{u} \, du \), which simplifies significantly.