Problem 34 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(\frac{d N}{d t}=3 N,\) where \(N=3.5\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \(N(t) = 3.5e^{3t}\). Substitution verifies the solution.

Step 1: Solve the differential equation

The given differential equation is \(\frac{d N}{d t} = 3N\). This is a first-order linear differential equation and can be solved using the method of separation of variables. To separate variables, rewrite as \(\frac{dN}{N} = 3 \, dt\).

Step 2: Integrate both sides

Integrate both sides of the equation: \(\int \frac{1}{N} \, dN = \int 3 \, dt\). The left side becomes \(\ln |N|\), and the right side becomes \(3t + C\), where \(C\) is the constant of integration. This gives us \(\ln |N| = 3t + C\).

Step 3: Solve for N

Exponentiate both sides to solve for \(N\): \(|N| = e^{3t + C}\). This simplifies to \(N = Ce^{3t}\), where \(C = e^C'\) is a new constant. Thus, \(N = Ce^{3t}\) is the general solution.

Step 4: Apply the initial condition to find particular solution

Use the initial condition \(N = 3.5\) when \(t = 0\) to find \(C\). Substitute into the general solution: \(3.5 = Ce^{0}\). Since \(e^{0} = 1\), \(C = 3.5\). The particular solution is \(N(t) = 3.5e^{3t}\).

Step 5: Verify the solution by substitution

Substitute \(N(t) = 3.5e^{3t}\) back into the original differential equation. Calculate \(\frac{dN}{dt}\) from the particular solution: \(\frac{dN}{dt} = \frac{d}{dt}(3.5e^{3t}) = 3.5 \times 3e^{3t} = 10.5e^{3t}\). The original equation \(\frac{dN}{dt} = 3N\) becomes \(10.5e^{3t} = 3 \times 3.5e^{3t}\), which simplifies to \(10.5e^{3t} = 10.5e^{3t}\), confirming that the particular solution satisfies the differential equation.

Key Concepts

Initial Condition

When solving differential equations, initial conditions play a crucial role in finding specific solutions. An initial condition provides a specific set of values at a certain point, often when time, \( t \), equals zero. This information helps us find unique solutions that meet those conditions. In our exercise, we're given \( N = 3.5 \) when \( t = 0 \). Meaning, at the start (time zero), the value of \( N \) is 3.5.
This is like setting a starting point for your solution. Without initial conditions, you could only find a general solution, which involves unknown constants. Initial conditions help us determine the value of these constants. Essentially, they pin down the answer to one specific solution out of many potential ones.

Separation of Variables

Separation of variables is a method to solve differential equations, used when you can separate the variables into two sides of the equation. The goal is to isolate all terms involving one variable on one side, and everything else on the other side.
In our exercise, we start with \( \frac{dN}{dt} = 3N \). By rearranging, we have \( \frac{dN}{N} = 3 \, dt \). This step allows us to work with each variable separately: \( N \) on one side and \( t \) on the other.

• This method is beneficial because it simplifies complex equations into two parts that can be integrated independently.
• Once the variables are separated, the next step is integration on both sides to solve the equation further.

Separation of variables is handy for equations that fit the criteria, making the solving process much more manageable.

Integration

The integration step is where we apply calculus to find the antiderivatives of each side of the separated equation. In this exercise, once the variables are separated, we integrate both sides:
\[ \int \frac{1}{N} \, dN = \int 3 \, dt \]
This gives us \( \ln |N| = 3t + C \), where \( C \) is the constant of integration. This constant is crucial because it represents all possible vertical shifts in the graph of the solution.

• Integration transforms a differential equation into an algebraic equation.
• Remember that whenever we integrate, we add an integration constant, represented by \( C \).

This constant will later be determined using the initial condition provided in the problem. This step is crucial for finding the general solution and moving towards the particular solution.

Particular Solution

The particular solution involves applying the initial condition to the general solution to find specific values for any constants. After integration, our general solution is \( N = Ce^{3t} \).

Now, using the initial condition \( N = 3.5 \) when \( t = 0 \), we can solve for \( C \). Plugging into the general solution, we find:

\[ 3.5 = Ce^0 \]

Since \( e^0 = 1 \), this simplifies to \( C = 3.5 \), making our particular solution \( N(t) = 3.5e^{3t} \).
The particular solution lets us describe the system's behavior specifically due to the initial condition. It's like finding the one path the system takes within all possible outcomes specified by the general solution.