Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 2: Problem 26
\(y^{\prime}+y \tan t=\cos t, y(0)=0\)
Short Answer
Expert verified
The solution is \( y = t \cos t \).
Step by step solution
01
Identify the type of differential equation
The given differential equation is a first-order linear ordinary differential equation of the form: \[ y' + P(t)y = Q(t) \] where \( P(t) = \tan t \) and \( Q(t) = \cos t \).
02
Find the integrating factor
The integrating factor \( \mu(t) \) is given by: \[ \mu(t) = e^{\int P(t) \, dt} \] Substituting \( P(t) = \tan t \), we get: \[ \mu(t) = e^{\int \tan t \, dt} \] The integral of \( \tan t \) is \( -\ln |\cos t| \), hence: \[ \mu(t) = e^{\ln |\sec t|} = |\sec t| \] Since \( \cos t \geq 0 \) in our domain, we simplify it to: \[ \mu(t) = \sec t \]
03
Multiply the differential equation by the integrating factor
Multiply both sides of the differential equation by \( \mu(t) = \sec t \): \[ \sec t \, y' + y \sec t \, \tan t = \sec t \cos t \] Simplifying the right hand side: \[ \sec t \, y' + y \sec t \, \tan t = y' \sec t + y \bigg( \frac{\sin t}{\cos t} \cdot \frac{1}{\cos t} \bigg) = y' \sec t + y \tan t \sec t := y' \sec t + y \sec t \tan t = 1 \]
04
Recognize the left side as the derivative of a product
The left-hand side of the equation now resembles the derivative of the product \( y \sec t \): \[ \frac{d}{dt} \big[ y \sec t \big] = 1 \]
05
Integrate both sides of the equation
Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt} \big[ y \sec t \big] \, dt = \int 1 \, dt \] The left-hand side simplifies to \( y \sec t \), leading to: \[ y \sec t = t + C \]
06
Solve for the arbitrary constant using the initial condition
Use the initial condition \( y(0) = 0 \) to find the constant \( C \): At \( t = 0 \), \( y(0) = 0 \): \( 0 \cdot \sec 0 = 0 + C \) Simplifying, since \( \sec 0 = 1 \): \[ 0 = C \] Thus, \( C = 0 \)
07
Write the final solution
Substitute \( C = 0 \) back into the equation: \[ y \sec t = t \] Solving for \( y \), we get: \[ y = t \cos t \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
To solve a first-order linear ordinary differential equation (ODE) like the one given, we need a special tool called the integrating factor. The general form of a first-order linear ODE is: \[ y' + P(t)y = Q(t) \] Here, \( P(t) \) and \( Q(t) \) are functions of \( t \). The integrating factor, \( \mu(t) \), helps transform the ODE into an easier form that can be integrated directly.
The integrating factor \( \mu(t) \) is computed as follows:
Let鈥檚 apply this to our provided equation: \( y' + y \tan t = \cos t \). Here, \( P(t) = \tan t \).
The integrating factor is: \[ \mu(t) = e^{\int \tan t \, dt} \] The integral of \( \tan t \) is \( -\ln |\cos t| \), so: \[ \mu(t) = e^{\ln |\sec t|} = |\sec t| \] Since for our domain \( \cos t \geq 0 \), we get: \( \mu(t) = \sec t \).
Multiplying the entire differential equation by \( \sec t \) facilitates solving it in the next steps.
The integrating factor \( \mu(t) \) is computed as follows:
- First, find \( P(t) \) from your ODE.
- Then, solve \( \mu(t) = e^{\int P(t) \, dt} \).
Let鈥檚 apply this to our provided equation: \( y' + y \tan t = \cos t \). Here, \( P(t) = \tan t \).
The integrating factor is: \[ \mu(t) = e^{\int \tan t \, dt} \] The integral of \( \tan t \) is \( -\ln |\cos t| \), so: \[ \mu(t) = e^{\ln |\sec t|} = |\sec t| \] Since for our domain \( \cos t \geq 0 \), we get: \( \mu(t) = \sec t \).
Multiplying the entire differential equation by \( \sec t \) facilitates solving it in the next steps.
Initial Condition
When solving differential equations, an initial condition is often provided. It specifies the value of the function at a particular point, and is crucial for finding the exact solution. For our equation, the initial condition is: \( y(0) = 0 \).
This means that when \( t = 0 \), the value of \( y \) is 0. We use this to find the constant of integration after integrating both sides of the equation.
Let's apply the initial condition to our solution process: First, we reached: \( y \sec t = t + C \).
Replacing \( t \) with 0, we find: \( y(0) = 0 \Rightarrow 0 \sec 0 = 0 + C \). Hence, \( C = 0 \).
This allows us to write the final solution as: \( y \sec t = t \) and therefore: \( y = t \cos t \). Applying the initial condition ensures our solution is unique and accurate for the given problem.
This means that when \( t = 0 \), the value of \( y \) is 0. We use this to find the constant of integration after integrating both sides of the equation.
Let's apply the initial condition to our solution process: First, we reached: \( y \sec t = t + C \).
Replacing \( t \) with 0, we find: \( y(0) = 0 \Rightarrow 0 \sec 0 = 0 + C \). Hence, \( C = 0 \).
This allows us to write the final solution as: \( y \sec t = t \) and therefore: \( y = t \cos t \). Applying the initial condition ensures our solution is unique and accurate for the given problem.
Solving Differential Equations
Solving a first-order linear ordinary differential equation entails several steps:
For our equation \( y' + y \tan t = \cos t \), we follow these steps:
1. Identify that it is a linear ODE with \( P(t) = \tan t \) and \( Q(t) = \cos t \).
2. Compute the integrating factor \( \mu(t) = \sec t \).
3. Multiply the equation by \( \sec t \): \( \sec t \cdot y' + y \tan t \sec t = 1 \).
4. Recognize the left side as \( \frac{d}{dt}(y \sec t) \).
5. Integrate both sides: \[ \int \frac{d}{dt}(y \sec t) \, dt = \int 1 \, dt \], leading to: \( y \sec t = t + C \).
6. Use the initial condition \( y(0) = 0 \) to find \( C = 0 \).
Thus, the final solution is: \( y = t \cos t \). Following a structured process helps ensure accuracy.
- Identify the form of the equation.
- Compute the integrating factor.
- Multiply the entire equation by the integrating factor.
- Rewrite the left side as the derivative of a product.
- Integrate both sides.
- Apply the initial condition to find any constants.
For our equation \( y' + y \tan t = \cos t \), we follow these steps:
1. Identify that it is a linear ODE with \( P(t) = \tan t \) and \( Q(t) = \cos t \).
2. Compute the integrating factor \( \mu(t) = \sec t \).
3. Multiply the equation by \( \sec t \): \( \sec t \cdot y' + y \tan t \sec t = 1 \).
4. Recognize the left side as \( \frac{d}{dt}(y \sec t) \).
5. Integrate both sides: \[ \int \frac{d}{dt}(y \sec t) \, dt = \int 1 \, dt \], leading to: \( y \sec t = t + C \).
6. Use the initial condition \( y(0) = 0 \) to find \( C = 0 \).
Thus, the final solution is: \( y = t \cos t \). Following a structured process helps ensure accuracy.
