Problem 63

Question

In Problems 59-72, solve the initial-value problem. $$ \frac{d N}{d t}=\frac{1}{t}, \text { for } t \geq 1 \text { with } N(1)=10 $$

Step-by-Step Solution

Verified

Answer

The solution is $ N(t) = \ln |t| + 10 $.

1Step 1: Identify the Differential Equation

The given differential equation is $ \frac{dN}{dt} = \frac{1}{t} $. This is a first-order differential equation.

2Step 2: 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 $.

3Step 3: 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 $.

4Step 4: 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 $.

5Step 5: Write the Solution

Substitute $ C = 10 $ back into the equation for $ N $, resulting in $ N(t) = \ln |t| + 10 $.

Key Concepts

First-Order Differential EquationSeparation of VariablesIntegrationInitial Condition

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 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 $.

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.

Use this method particularly when the equation can be easily rearranged so that each side involves a different variable.

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 $.

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.

Integration transforms a rate of change (derivative) back into the original function, providing the solution to our equation. This allows us to apply initial conditions to find the specific solution that fits the problem context.

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 $.

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 $.

The initial condition is essential to tailor the general solution, $ N(t) = \ln |t| + 10 $, to fit the problem's specific requirements. It essentially roots the solution in reality, corresponding to the actual initial state of the system described by the differential equation.

Problem 63

Problem 64

Other exercises in this chapter

Problem 62

In Problems 59-72, solve the initial-value problem. $$ \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}, \text { for } x \geq 1 \text { with } y=3 \text { when } x=4 $$

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Problem 63

For $p>0$, determine the values of $p$ for which the following limit is either 1 or $\infty$ or a constant that is neither 1 nor $\infty$ : $$ \lim _{x

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Problem 64

Show that $$ \lim _{x \rightarrow \infty} x^{p} e^{-x}=0 $$ for any positive number $p .$ Graph $f(x)=x^{p} e^{-x}, x>0$, for $p=1 / 2,1$, and $2 .$ Sin

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Problem 64

In Problems 59-72, solve the initial-value problem. $$ \frac{d N}{d t}=\frac{t}{t+2}, \text { for } t \geq 0 \text { with } N(0)=2 $$

View solution