Problem 41 - Advanced Engineering Mathematics

Question

We want the general solution to be \(y=3 x-5+c e^{-x}\). (Rather than \(e^{-x}\), any function that approaches 0 as \(x \rightarrow \infty\) could be used.) Differentiating we get \\[ y^{\prime}=3-c e^{-x}=3-(y-3 x+5)=-y+3 x-2 \\] so the differential equation \(y^{\prime}+y=3 x-2\) has solutions asymptotic to the line \(y=3 x-5\).

Step-by-Step Solution

Answer

The function \(y=3x-5+ce^{-x}\) satisfies the differential equation and approaches \(y=3x-5\) as \(x \rightarrow \infty\).

Step 1: Understand the Problem

We need to verify that the given function \(y = 3x - 5 + ce^{-x}\) is a solution to the differential equation \(y' + y = 3x - 2\). The problem also mentions that the solutions are asymptotic to the line \(y = 3x - 5\).

Step 2: Differentiate the Given Function

First, let's differentiate the given function with respect to \(x\):The function is \(y = 3x - 5 + ce^{-x}\). Its derivative \(y'\) is computed as follows:\[y' = \frac{d}{dx}(3x) - \frac{d}{dx}(5) + \frac{d}{dx}(ce^{-x}) = 3 + 0 - ce^{-x} = 3 - ce^{-x}.\]

Step 3: Substitute into the Differential Equation

Substitute \(y = 3x - 5 + ce^{-x}\) and \(y' = 3 - ce^{-x}\) into the differential equation \(y' + y = 3x - 2\):- Left-hand side: \(y' + y = (3 - ce^{-x}) + (3x - 5 + ce^{-x})\).- Simplifies to: \(3 + 3x - 5\).- Further simplifies to: \(3x - 2\).This matches the right-hand side of the differential equation, confirming that the function satisfies \(y' + y = 3x - 2\).

Step 4: Verify Asymptotic Behavior

As \(x \to \infty\), the term \(ce^{-x}\) approaches zero, making the function approach \(y = 3x - 5\). Hence, solutions asymptotically approach the line \(y = 3x - 5\) as required.

Key Concepts

Asymptotic Solutions

In the context of differential equations, an asymptotic solution describes a function's behavior as one or more variables approach a limit, often infinity. For the problem at hand, the asymptotic behavior refers to the function approaching a straight line as \(x \to \infty\).
\[
\begin{align*}
y &= 3x - 5 + ce^{-x}\
y' &= 3 - ce^{-x}\
\end{align*}
\]
The term \(ce^{-x}\) is crucial as it approaches zero when \(x\) grows larger. Consequently, the overall function \(y = 3x - 5 + ce^{-x}\) approaches the line \(y = 3x - 5\) for large values of \(x\).
This property is essential when analyzing the long-term behavior of a solution. In practical terms, it implies that over time, or over a large interval, the impact of the term \(ce^{-x}\) diminishes, and the function closely follows the line \(y = 3x - 5\).
Having a broader understanding of asymptotic solutions can help predict the function's behavior without solving the differential equation explicitly.

General Solution

The general solution of a differential equation is a formula that provides all possible solutions. It typically includes arbitrary constants, which can be adjusted to fit specific initial conditions or boundaries.
In our example, the equation \(y' + y = 3x - 2\) has a general solution \(y = 3x - 5 + ce^{-x}\).

• The term \(3x - 5\) represents a particular (or specific) solution. It satisfies the differential equation without contributing to any transient behavior.
• The term \(ce^{-x}\) incorporates an arbitrary constant \(c\). This component is crucial for adjusting the function to meet any specified conditions.

The beauty of the general solution is that it accounts for all possible particular solutions that the differential equation might have.
Changing the value of \(c\) gives us various solutions, enabling accommodation of different scenarios or parameters inside the same framework.
This flexibility is vital in fields like physics and engineering, where initial conditions can vary, requiring customized solutions to a problem.

Differentiation

Differentiation is a fundamental concept in calculus. It's the process of finding the derivative of a function, which tells us how the function changes with respect to one of its variables.
In our problem, differentiation helps determine how the function \(y = 3x - 5 + ce^{-x}\) changes with respect to \(x\).
The derivative, \(y' = 3 - ce^{-x}\), shows:

• The term \(3\) is the constant rate of change associated with the linear component \(3x\).
• The term \(-ce^{-x}\) demonstrates how the exponential term contributes to changes in the function. As \(x\) increases, this part has less impact because \(e^{-x}\) approaches zero.

Understanding differentiation is critical because it connects to numerous applications.

For instance, in physics, the derivative of a position function concerning time yields velocity. Similarly, in our problem, it helped verify that the function fits within the differential equation \(y' + y = 3x - 2\).
As such, differentiation provides profound insights not only into the behavior of functions but also into the dynamic systems they represent.