Problem 23 - 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. $$ x \frac{d y}{d x}+(3 x+1) y=e^{-3 x} $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{1}{e^{3x}} + \frac{C}{xe^{3x}} \) over the interval \( (0, \infty) \). The transient term is \( \frac{C}{xe^{3x}} \).

Step 1: Identify the type of differential equation

The given equation, \( x \frac{d y}{d x} + (3x + 1)y = e^{-3x} \), is a first-order linear differential equation of the form \( a(x)y' + b(x)y = c(x) \). In this case, \( a(x) = x \), \( b(x) = 3x + 1 \), and \( c(x) = e^{-3x} \).

Step 2: Divide by the leading coefficient

To simplify the equation and separate the variable, divide through by \( x \) (assuming \( x eq 0 \)) to get the standard form: \( \frac{d y}{d x} + \frac{3x + 1}{x} y = \frac{e^{-3x}}{x} \).

Step 3: Compute the integrating factor

The integrating factor \( \mu(x) \) is calculated as \( e^{\int \frac{3x + 1}{x} \, dx} = e^{\int (3 + \frac{1}{x}) \, dx} = e^{3x + \ln|x|} = x e^{3x} \).

Step 4: Multiply the equation by the integrating factor

Multiply every term in the differential equation by the integrating factor \( xe^{3x} \): \( (xe^{3x}) \cdot \frac{d y}{d x} + (xe^{3x}) \cdot \frac{3x + 1}{x} y = (xe^{3x}) \cdot \frac{e^{-3x}}{x} \).

Step 5: Recognize the left-hand side as a derivative

The left-hand side simplifies to the derivative of a product: \( \frac{d}{dx} (ye^{3x}x) = e^{3x} \). Set this equal to the right-hand side to get: \( \frac{d}{dx} (yxe^{3x}) = 1 \).

Step 6: Integrate both sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx} (yxe^{3x}) \, dx = \int 1 \, dx \). This yields: \( yxe^{3x} = x + C \) for some constant \( C \).

Step 7: Solve for \( y \)

To find the general solution, solve for \( y \):\( y = \frac{x + C}{x e^{3x}} = \frac{1}{e^{3x}} + \frac{C}{xe^{3x}} \).

Step 8: Determine the interval of definition

The general solution is defined for \( x > 0 \) since we divided by \( x \) initially. Hence, the largest interval over which \( y \) is defined is \( (0, \infty) \).

Step 9: Identify transient terms

The transient term in the general solution is \( \frac{C}{xe^{3x}} \) because these terms vanish as \( x \to \infty \).

Key Concepts

Integrating Factor

First-order linear differential equations, such as the one given in the exercise, often utilize the concept of an integrating factor to simplify and solve them. To make the differential equation solvable, we need to find a function that can multiply the entire equation and convert the left-hand side into an exact derivative of a product.

The general approach involves identifying the coefficient of the dependent variable term in the standard form. This coefficient affects how we calculate the integrating factor. In our given differential equation, after converting it to the standard form:

• The term \( y' \) has a coefficient transformed into \( \frac{3x + 1}{x} \),

which also serves as the expression to be integrated. The integrating factor, \( \mu(x) \), is derived from:

• \( e^{\int (3 + \frac{1}{x}) dx} = e^{3x + \ln|x|} \) .

After calculating, it simplifies to \( xe^{3x} \). By multiplying the entire differential equation by this integrating factor, the equation transforms, allowing the left-hand side to be easily expressed as a single derivative. Now, solving the equation becomes a matter of integration.

General Solution

The general solution of a differential equation represents a family of possible solutions, determined by an arbitrary constant. In the context of the given exercise, we aimed to find a function \( y(x) \) that meets the equation's constraints under specific terms.

Once the integrating factor \( xe^{3x} \) is applied, the left side of the equation becomes the derivative of \( yxe^{3x} \). This is due to the technique that allows us to convert into a derivative form:

• \( \frac{d}{dx} (yxe^{3x}) = 1 \).

Integrating both sides with respect to \( x \) yields \( yxe^{3x} = x + C \), where \( C \) is the integration constant.

By solving for \( y \), we arrive at the general solution:

• \( y = \frac{x + C}{xe^{3x}} = \frac{1}{e^{3x}} + \frac{C}{xe^{3x}} \) .

This solution is defined over the interval \( x > 0 \), as inferred from the steps involved in deriving and integrating the equation.

Transient Terms

Transient terms describe parts of a solution that diminish over time or as the variable approaches infinity. In the general solution found earlier:

• \( y = \frac{1}{e^{3x}} + \frac{C}{xe^{3x}} \) ,

the component \( \frac{C}{xe^{3x}} \) is identified as a transient term.

Transient terms are particularly important in applications such as physics or engineering because they indicate temporary behaviors that eventually fade or stabilize. As \( x \) increases, \( \frac{C}{xe^{3x}} \) tends towards zero, implying this term's effect becomes negligible at large \( x \) values.

This allows the solution to stabilize to a more permanent state, often reflected by the remaining non-transient part \( \frac{1}{e^{3x}} \). Recognizing transient terms helps in understanding the dynamic evolution of the system described by the differential equation.