Problem 18 - Advanced Engineering Mathematics

Question

From \(\frac{1}{N} d N=\left(t e^{t+2}-1\right) d t\) we obtain \(\ln |N|=t e^{t+2}-e^{t+2}-t+c\) or \(N=c_{1} e^{t e^{t+2}-e^{t+2}-t}\).

Step-by-Step Solution

Answer

\( N = c_1 e^{t e^{t+2} - e^{t+2} - t} \).

Step 1: Introduce the Given Differential Equation

We start with the given equation \( \frac{1}{N} \, dN = (t e^{t+2} - 1) \, dt \). This is a separable differential equation, allowing us to separate the variables \( N \) and \( t \).

Step 2: Separate Variables and Integrate

Separate the variables to get \( \int \frac{1}{N} \, dN = \int (t e^{t+2} - 1) \, dt \). Integrate both sides: \( \ln |N| = \int t e^{t+2} \, dt - \int 1 \, dt \).

Step 3: Solve the First Integral

To solve \( \int t e^{t+2} \, dt \), use integration by parts. Let \( u = t \) and \( dv = e^{t+2} \, dt \). Then \( du = dt \) and \( v = e^{t+2} \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 4: Apply Integration by Parts

Applying the formula: \( \int t e^{t+2} \, dt = t e^{t+2} - \int e^{t+2} \, dt \). The second integral is straightforward: \( \int e^{t+2} \, dt = e^{t+2} + C \), so \( \int t e^{t+2} \, dt = t e^{t+2} - e^{t+2} \).

Step 5: Solve the Second Integral

The second integral \( \int 1 \, dt = t \). Add the constant \( C \) for the indefinite integrals: \( \ln |N| = t e^{t+2} - e^{t+2} - t + C \).

Step 6: Solve for N

Exponentiate both sides to solve for \( N \): \( |N| = e^{t e^{t+2} - e^{t+2} - t + C} \). Simplify to \( N = c_1 e^{t e^{t+2} - e^{t+2} - t} \), where \( c_1 = e^C \) is an arbitrary constant.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals involving the product of two functions. It is especially handy when one of the functions easily integrates while the other simplifies when differentiated.
To apply integration by parts, we use the formula:

• \( \int u \, dv = uv - \int v \, du \)

Here, you choose one function as \( u \) to be differentiated, and the other as \( dv \) to be integrated.
For example, in the integral \( \int t e^{t+2} \, dt \), we select \( u = t \) and \( dv = e^{t+2} \, dt \). This choice simplifies the calculations as \( du = dt \) and \( v = e^{t+2} \), forming a straightforward integration.
The final calculation results in \( t e^{t+2} - e^{t+2} \), which is much simpler to handle in further computations.

Differential Equation Solutions

Differential equations involve derivatives and are used to describe various physical phenomena. Solving them often means finding a function that satisfies the given relationship between variables.
In the exercise, we are working with a separable differential equation, which allows us to rearrange terms so each variable occupies its own side of the equation. This makes it easier to integrate each part separately.
Starting with \( \frac{1}{N} \, dN = (t e^{t+2} - 1) \, dt \), we can separate and then integrate both sides:

• \( \int \frac{1}{N} \, dN = \ln |N| \)
• \( \int (t e^{t+2} - 1) \, dt \) involves using integration by parts for \( t e^{t+2} \) and a simple integral for \( -1 \).

Ultimately, solving this results in \( \ln |N| = t e^{t+2} - e^{t+2} - t + C \), where \( C \) is the constant of integration. Solving this expression involves exponentiating to isolate \( N \).
The solution to the differential equation, \( N = c_1 e^{t e^{t+2}-e^{t+2} - t} \), provides the general form of the function \( N(t) \).

Indefinite Integrals

Indefinite integrals represent families of functions and include a constant of integration, symbolized as \( C \), because their derivative yields the original function being integrated.
In our exercise, we encounter indefinite integrals such as:

• \( \int \frac{1}{N} \, dN = \ln |N| + C \)
• \( \int (t e^{t+2} - 1) \, dt \), which breaks into two separate integrals using integration by parts and basic rules.

The role of the constant \( C \) is crucial, as it accounts for the family of curves that represent all possible solutions given any initial conditions.
Integrating results in expressions that contain \( C \), reminding us that without specific boundary conditions or additional information, our solutions remain generalized.
Indefinite integrals thus offer flexibility but require further data to specify unique solutions.