Problem 19 - Advanced Engineering Mathematics

Question

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ (x+1) \frac{d y}{d x}+(x+2) y=2 x e^{-x} $$

Step-by-Step Solution

Answer

The general solution is \(y = \frac{x^2 + C}{e^x(x+1)}\), valid for \(x \neq -1\). The term \(x^2/e^x\) is transient.

Step 1: Identify the Type of Differential Equation

The given differential equation is \((x+1) \frac{dy}{dx} + (x+2)y = 2xe^{-x}\). Upon inspection, it is a first-order linear differential equation since it can be reorganized into the standard form \(P(x) \frac{dy}{dx} + Q(x)y = G(x)\).

Step 2: Rewrite in Standard Form

Divide every term by \(x+1\) to rewrite the equation into the standard form \( \frac{dy}{dx} + \frac{x+2}{x+1} y = \frac{2x}{x+1} e^{-x}\). This confirms the identification of the differential equation.

Step 3: Find the Integrating Factor

The integrating factor, \(\mu(x)\), is derived from \(e^{\int P(x)\, dx}\). Here, \(P(x) = \frac{x+2}{x+1}\), so: \[\mu(x) = e^{\int \frac{x+2}{x+1} \, dx} = e^{\int \left(1 + \frac{1}{x+1}\right) \, dx} = e^{x} e^{\ln|x+1|} = e^{x} |x+1|.\]

Step 4: Solve the Differential Equation

Multiply the entire differential equation by \(\mu(x) = e^{x}(x+1)\) to make it exact: \[e^{x}(x+1) \frac{dy}{dx} + e^{x}(x+2)y = 2xe^{-x}(x+1).\] Integrate both sides: \[ \frac{d}{dx}(e^{x}(x+1)y) = 2x. \] Therefore, \[e^{x}(x+1)y = \int 2x \, dx = x^2 + C.\] Solve for \(y\): \[ y = \frac{x^2 + C}{e^x(x+1)}.\]

Step 5: Determine the Interval of the Solution

The general solution has terms \(x+1\) and \(e^x\). The domain is where these are defined. The function \(\frac{1}{x+1}\) is defined for \(x eq -1\). The interval for \(x\) is therefore \((-\infty, -1) \cup (-1, \infty)\).

Step 6: Identify Transient Terms

Transient terms are those that disappear as \(x\) approaches \(\infty\) or \(-\infty\). In the general solution \(y = \frac{x^2 + C}{e^x(x+1)}\), as \(x \to \infty\), the exponential term \(e^x\) dominates, causing \(y\to 0\). The term \(x^2/(e^x)\) is indeed transient.

Key Concepts

General Solution

In solving a first-order linear differential equation like \((x+1) \frac{dy}{dx} + (x+2)y = 2xe^{-x}\), the ultimate goal is to find the general solution. A general solution is an expression that includes all possible solutions to the differential equation.
In this particular example, after applying the method of integrating factors, we find that the general solution is: \[ y = \frac{x^2 + C}{e^x(x+1)}. \]
This expression represents a family of curves, each defined by a different value of the constant \(C\). The constant is typically determined when initial conditions are available.

• The general solution incorporates the structure of the original differential equation.
• It reveals a comprehensive set of results without needing specific initial conditions.

Thus, the general solution is crucial because it captures the complete nature of the differential equation.

Integrating Factor

Solving first-order linear differential equations often involves finding an integrating factor. This factor makes the left-hand side of the differential equation into the derivative of a product, simplifying the integration process.
For this differential equation, the integrating factor \(\mu(x)\) is identified as:\[ \mu(x) = e^{\int \frac{x+2}{x+1} \, dx} = e^x |x+1|. \]
This factor is obtained by:

• Recognizing \(P(x)\) from the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).
• Computing the exponential of the integral of \(P(x)\).
• Utilizing properties of logarithms and exponentials.

With the integrating factor, the differential equation is made exact, facilitating the integration needed to find the solution. It's a powerful tool that transforms the problem into a more manageable form.

Interval of Definition

The interval of definition refers to the range over which the general solution of a differential equation is valid, based on the terms present in the solution.
For the solution \( y = \frac{x^2 + C}{e^x(x+1)} \), the critical consideration is the denominator \(x+1\), which becomes zero at \(x = -1\). Consequently, the solution is not defined where the denominator is zero. Therefore, the largest interval where the solution is defined is:\(( -\infty, -1) \cup (-1, \infty)\).

• Recognizes points of discontinuity or undefined terms.
• Acknowledges similar constraints from other parts of the equation (e.g., logarithmic terms).

Understanding this interval is essential as it tells us where the solutions are physically or mathematically meaningful.

Transient Terms

In analyzing differential equations, especially as \(x\) approaches extreme values (like \(\infty\) or \(-\infty\)), identifying transient terms within the general solution is important.
Transient terms are those that have diminishing effects over time. In this particular solution:\[ y = \frac{x^2 + C}{e^x(x+1)}, \]we observe that as \(x \to \infty\), the term \(x^2/e^x\) tends towards zero because of the overpowering growth rate of \(e^x\).

• The term is considered transient because it fades with increasing \(x\).
• Critical for predicting long-term behavior of the system described by the differential equation.

Transient terms essentially 'disappear,' leaving the system's steady state or perpetual behavior prominent.