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Magazine Chapter 5: Problem 63
In Problems 59-72, solve the initial-value problem. $$ \frac{d N}{d t}=\frac{1}{t}, \text { for } t \geq 1 \text { with } N(1)=10 $$
Short Answer
Expert verified
The solution is \( N(t) = \ln |t| + 10 \).
Step by step solution
01
Identify the Differential Equation
The given differential equation is \( \frac{dN}{dt} = \frac{1}{t} \). This is a first-order differential equation.
02
Separate Variables
We can manipulate the equation to get all terms with \( N \) on one side and all terms with \( t \) on the other side, resulting in \( dN = \frac{1}{t} dt \).
03
Integrate Both Sides
Integrate both sides to solve for \( N \). We integrate \( \int dN \) and \( \int \frac{1}{t} dt \), which yields \( N = \ln |t| + C \).
04
Apply the Initial Condition
Use the initial condition \( N(1) = 10 \). Substitute \( t = 1 \) and \( N = 10 \) into the equation: \( 10 = \ln |1| + C \). Since \( \ln |1| = 0 \), we find \( C = 10 \).
05
Write the Solution
Substitute \( C = 10 \) back into the equation for \( N \), resulting in \( N(t) = \ln |t| + 10 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equation
A first-order differential equation is a mathematical equation that relates a function with its first derivative. These equations often describe rates of change. They are called 'first-order' because they involve only the first derivative of the unknown function.
In our problem, the differential equation is given by \( \frac{dN}{dt} = \frac{1}{t} \). This tells us that the rate of change of the function \( N \) with respect to \( t \) is equal to \( \frac{1}{t} \). First-order differential equations can model many real-world processes, such as exponential decay, population growth, or cooling processes.
In our problem, the differential equation is given by \( \frac{dN}{dt} = \frac{1}{t} \). This tells us that the rate of change of the function \( N \) with respect to \( t \) is equal to \( \frac{1}{t} \). First-order differential equations can model many real-world processes, such as exponential decay, population growth, or cooling processes.
- In our equation, \( t \) is the independent variable, while \( N \) is the dependent variable.
- The equation does not contain higher derivatives, which simplifies analysis and solution techniques.
Separation of Variables
Separation of variables is a method used to solve first-order differential equations by expressing the equation in a way where each variable appears on its own side of the equation. This method often simplifies the integration process.
For the given problem, we manipulate the equation \( \frac{dN}{dt} = \frac{1}{t} \) to separate the variables. We achieve this by rearranging the equation to \( dN = \frac{1}{t} dt \).
For the given problem, we manipulate the equation \( \frac{dN}{dt} = \frac{1}{t} \) to separate the variables. We achieve this by rearranging the equation to \( dN = \frac{1}{t} dt \).
- Separating variables involves organizing the equation's terms so that all instances of one variable are on one side, and all instances of the other variable are on the opposite side.
- This allows us to integrate both sides independently, leading us closer to finding a solution for the unknown function.
Integration
Integration is a fundamental concept in calculus, used to solve differential equations. It involves finding a function whose derivative is the given function.
In the equation \( dN = \frac{1}{t} dt \), we integrate both sides to solve for \( N \).
In the equation \( dN = \frac{1}{t} dt \), we integrate both sides to solve for \( N \).
- On the left side, integrating \( \int dN \) gives us \( N \).
- On the right side, integrating \( \int \frac{1}{t} dt \) results in \( \ln |t| + C \), where \( C \) is the constant of integration. The constant is crucial as it represents the general solution set.
Initial Condition
An initial condition provides specific information that allows us to find a unique solution out of a family of possible solutions.
In our problem, the initial condition is given by \( N(1) = 10 \). This tells us the value of the function \( N \) when \( t = 1 \).
In our problem, the initial condition is given by \( N(1) = 10 \). This tells us the value of the function \( N \) when \( t = 1 \).
- By substituting \( t = 1 \) and \( N = 10 \) into the integrated equation \( N = \ln |t| + C \), we solve for \( C \).
- In this case, when \( t = 1 \), \( \ln |1| = 0 \). Therefore, \( 10 = 0 + C \), giving us \( C = 10 \).
