Chapter 5: Problem 63
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 \( N(t) = \ln t + 10 \) for \( t \geq 1 \).
Step by step solution
01
Understand the Differential Equation
We are given the differential equation \( \frac{d N}{d t} = \frac{1}{t} \). This is a first-order separable differential equation, and our goal is to find \( N(t) \) that satisfies this equation and the initial condition \( N(1) = 10 \).
02
Set Up the Integration
To solve \( \frac{d N}{d t} = \frac{1}{t} \), we integrate both sides with respect to \( t \). This gives us \( N(t) = \int \frac{1}{t} \, dt \).
03
Integrate the Function
Perform the integration: \( \int \frac{1}{t} \, dt = \ln |t| + C \), where \( C \) is the integration constant. Thus, \( N(t) = \ln |t| + C \).
04
Apply the Initial Condition
Use the initial condition \( N(1) = 10 \) to find \( C \). Substitute \( t = 1 \) into \( N(t) = \ln |t| + C \) to get \( 10 = \ln 1 + C \). Since \( \ln 1 = 0 \), we find that \( C = 10 \).
05
Write the Solution
The solution to the initial-value problem is \( N(t) = \ln |t| + 10 \). Since \( t \geq 1 \), we can simplify this to \( N(t) = \ln t + 10 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Differential Equation
A separable differential equation is a type of first-order differential equation that can be written in the form of `\( \frac{dy}{dx} = g(x)h(y) \)`. The beauty of this form lies in its ability to be split into two parts, one involving only \( x \) and the other only \( y \), which we can then integrate independently. This type of equation is common in real-world applications due to its simplicity and the straightforward strategy for solving it.
In the case of our differential equation, `\( \frac{d N}{d t} = \frac{1}{t} \)`, the equation is already separable. We isolate the variables so that we have all the \( N \) terms on one side and all the \( t \) terms on the other. This step essentially transforms the equation into an integrable form:
- Move \( dt \) to the other side: `\( dN = \frac{1}{t}dt \)`.
- Both sides are now ready to be integrated independently.
This separation of variables is crucial as it allows us to rewrite the equation in a way that is conducive to integration, leading us to the solution of the differential equation.
In the case of our differential equation, `\( \frac{d N}{d t} = \frac{1}{t} \)`, the equation is already separable. We isolate the variables so that we have all the \( N \) terms on one side and all the \( t \) terms on the other. This step essentially transforms the equation into an integrable form:
- Move \( dt \) to the other side: `\( dN = \frac{1}{t}dt \)`.
- Both sides are now ready to be integrated independently.
This separation of variables is crucial as it allows us to rewrite the equation in a way that is conducive to integration, leading us to the solution of the differential equation.
Integration Constant
When you solve a separable differential equation by integrating, it’s important to remember the integration constant. This constant arises because indefinite integration can have multiple solutions, each differing by a constant term. Whenever we integrate a function, we can add any constant value, which is why we include +\( C \) in our integration results.
In the given problem, after integrating \( \int \frac{1}{t} \, dt \), we get \( \ln |t| + C \). Here:
- \( C \) is an unknown constant of integration.
- We initially don't know its exact value until we apply the initial condition, which helps us solve for \( C \).
In practice, the integration constant accounts for the specific solution to our initial-value problem. Without it, our solution would represent a family of infinitely many potential curves rather than the specific curve that satisfies the given initial condition.
In the given problem, after integrating \( \int \frac{1}{t} \, dt \), we get \( \ln |t| + C \). Here:
- \( C \) is an unknown constant of integration.
- We initially don't know its exact value until we apply the initial condition, which helps us solve for \( C \).
In practice, the integration constant accounts for the specific solution to our initial-value problem. Without it, our solution would represent a family of infinitely many potential curves rather than the specific curve that satisfies the given initial condition.
Initial Condition
An initial condition, in this context, is a piece of information that allows us to find the specific solution to a differential equation that would pass through a given point on its graph. It can be thought of as an anchor point.
In the problem, the initial condition is given as \( N(1) = 10 \). This condition is used to determine the value of the integration constant \( C \).
Upon reaching the solution \( N(t) = \ln |t| + C \), applying the initial condition involves:
- Substituting \( t = 1 \) into the equation:
\( N(1) = \ln |1| + C = 10 \)
- Since \( \ln 1 = 0 \), the equation simplifies to \( C = 10 \).
Thus, the initial condition allows us to pin down our particular solution to \( N(t) = \ln t + 10 \). This means that as \( t \) increases starting from \( t = 1 \), the function behavior is exactly determined by this specific formula, perfectly matching the given initial data.
In the problem, the initial condition is given as \( N(1) = 10 \). This condition is used to determine the value of the integration constant \( C \).
Upon reaching the solution \( N(t) = \ln |t| + C \), applying the initial condition involves:
- Substituting \( t = 1 \) into the equation:
\( N(1) = \ln |1| + C = 10 \)
- Since \( \ln 1 = 0 \), the equation simplifies to \( C = 10 \).
Thus, the initial condition allows us to pin down our particular solution to \( N(t) = \ln t + 10 \). This means that as \( t \) increases starting from \( t = 1 \), the function behavior is exactly determined by this specific formula, perfectly matching the given initial data.