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Magazine Chapter 12: Problem 12
Solve the initial value problem. \(y^{\prime}(t)+y(t)=1\), with \(y(0)=2\)
Short Answer
Expert verified
The solution is \( y(t) = 1 + e^{-t} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is linear and first-order. It is in the form \( y'(t) + p(t) y(t) = g(t) \). Here, \( p(t) = 1 \) and \( g(t) = 1 \).
02
Find the Integrating Factor
The integrating factor \( \mu(t) \) is found using the formula \( \mu(t) = e^{\int p(t) \, dt} \). Given \( p(t) = 1 \), we have \( \mu(t) = e^{\int 1 \, dt} = e^t \).
03
Multiply the Equation by the Integrating Factor
Multiply the entire differential equation \( y'(t) + y(t) = 1 \) by \( e^t \) to get \( e^t y'(t) + e^t y(t) = e^t \).
04
Recognize the Left Hand Side as a Derivative
The left-hand side of the equation \( e^t y'(t) + e^t y(t) \) can be written as the derivative of a product: \( \frac{d}{dt} [e^t y(t)] \). The equation is now \( \frac{d}{dt} [e^t y(t)] = e^t \).
05
Integrate Both Sides
Integrate both sides with respect to \( t \). The integral on the left is \( \int \frac{d}{dt} [e^t y(t)] \, dt = e^t y(t) \), and the integral on the right is \( \int e^t \, dt = e^t + C \), where \( C \) is the constant of integration.
06
Solve for \( y(t) \)
From the integrated form \( e^t y(t) = e^t + C \), divide both sides by \( e^t \) to isolate \( y(t) \): \( y(t) = 1 + Ce^{-t} \).
07
Apply the Initial Condition
Use the initial condition \( y(0) = 2 \) to solve for \( C \). Substitute \( t = 0 \) and \( y(0) = 2 \) into \( 1 + Ce^{-0} = 2 \), resulting in \( 1 + C = 2 \). Therefore, \( C = 1 \).
08
Write the Final Solution
Substitute \( C = 1 \) back into the expression for \( y(t) \) to get the final solution: \( y(t) = 1 + e^{-t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation along with a specified value at a starting point. For example, our problem with the equation \(y'(t) + y(t) = 1\) and the initial condition \(y(0) = 2\). The main goal is to find a function \(y(t)\) that not only satisfies the differential equation but also meets the initial condition.
This is how it works:
This is how it works:
- Start with a differential equation.
- Add an initial condition specifying the value of the solution at a particular time, usually \(t = 0\).
- Use methods like integrating factors to solve for \(y(t)\).
Integrating Factor
The Integrating Factor is a technique used to solve linear first-order differential equations. It simplifies these equations by turning them into a form that can be directly integrated.
To find an integrating factor, \(\mu(t)\), you use the formula \(\mu(t) = e^{\int p(t) \, dt}\).
Here's how it works:
To find an integrating factor, \(\mu(t)\), you use the formula \(\mu(t) = e^{\int p(t) \, dt}\).
Here's how it works:
- Identify \(p(t)\) from the differential equation.
- Compute \(\mu(t)\), which for our example becomes \(e^t\) since \(p(t) = 1\).
- Multiply the entire equation by \(\mu(t)\), transforming it into a more manageable form.
Linear First-Order Differential Equations
A Linear First-Order Differential Equation has the general form \(y'(t) + p(t) y(t) = g(t)\). It's 'linear' because the terms involve only powers of \(y\) or its derivatives.
Key steps to solve them include:
Key steps to solve them include:
- Recognize the equation is linear and first-order.
- Use the integrating factor method to simplify.
- Integrate to find the general solution.
