Problem 8 - 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. $$ y^{\prime}=2 y+x^{2}+5 $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{x^2}{2} + \frac{5}{2} + Ce^{-2x} \). Defined on \((-\infty, \infty)\), with a transient term \(Ce^{-2x}\).

Step 1: Identify the Differential Equation Type

The given differential equation is \( y' = 2y + x^2 + 5 \). This is a first-order linear differential equation of the form \( y' = ay + f(x) \), where \( a = 2 \) and \( f(x) = x^2 + 5 \).

Step 2: Find Integrating Factor

To solve the differential equation, we find an integrating factor \( \mu(x) = e^{\int a \, dx} = e^{\int 2 \, dx} = e^{2x} \). This will help us solve the equation easily.

Step 3: Multiply Equation by Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2x} \): \[ e^{2x}y' = 2e^{2x}y + e^{2x}(x^2 + 5) \]

Step 4: Recognize as Exact Differential

Notice that the left-hand side of the equation is the derivative of \( e^{2x}y \). Thus, the equation becomes:\[ \frac{d}{dx}(e^{2x}y) = e^{2x}(x^2 + 5) \]

Step 5: Integrate Both Sides

Integrate both sides to find \( y \):\[ e^{2x}y = \int e^{2x}(x^2 + 5) \, dx \]Integration by parts might be necessary here for \( \int e^{2x} x^2 \, dx \). After integration, we solve for \( y \).

Step 6: Solve for y

After integrating, we find the expression for \( e^{2x}y \) and solving for \( y \) yields:\[ y = e^{-2x}\left(\frac{x^2}{2}e^{2x} + rac{5}{2}e^{2x} + C\right) \]Simplifying gives:\[ y = \frac{x^2}{2} + \frac{5}{2} + Ce^{-2x} \]

Step 7: Identify the Interval of Solution

Since there are no restrictions on \( x \) in \( f(x) = x^2 + 5 \) and exponential functions are defined for all real numbers, the largest interval over which this solution is defined is \( (-\infty, \infty) \).

Step 8: Determine Transient Terms

Transient terms in a solution usually involve the exponential terms that decay to zero as \( x \to \infty \). Here, \( Ce^{-2x} \) is a transient term because it decays as \( x \to \infty \).

Key Concepts

Integrating Factor

In first-order linear differential equations, the integrating factor is a special tool that helps simplify and solve these equations. Consider the standard form of a first-order linear differential equation:

• \( y' = a y + f(x) \).

To solve such equations, we use an integrating factor, \( \mu(x) \), calculated as follows:

• First, find the constant \( a \) in the equation.
• Then, calculate the integrating factor as \( \mu(x) = e^{\int a \, dx} \).

In our problem, the constant \( a \) is 2, so the integrating factor becomes \( \mu(x) = e^{2x} \).

This integrating factor helps transform the equation into a form that is easier to work with. By multiplying the original equation by the integrating factor, you set the stage for solving the differential equation.

The main takeaway is that finding an integrating factor is crucial because it turns the differential equation into something that can be straightforwardly integrated.

General Solution

The general solution of a differential equation provides a formula that includes all possible solutions of the differential equation.

In the context of first-order linear differential equations, once you find the integrating factor and apply it to the equation, the process of integration gives the general solution.

• For instance, our solution began by multiplying the whole equation by \( e^{2x} \), the integrating factor.
• The equation transformed into the exact differential form \( \frac{d}{dx}(e^{2x}y) = e^{2x}(x^2 + 5) \).
• Integrating both sides, and solving for \( y \), yields:
• \( y = \frac{x^2}{2} + \frac{5}{2} + Ce^{-2x} \).

The undesignated constant \( C \) represents the constant of integration, expressing that this is a general solution covering all possible specific solutions.

Ultimately, the general solution describes how the function\( y \) behaves under the given differential equation, allowing for constant flexibility \( C \) in the function's behavior.

Transient Term

Transient terms in a differential equation's solution are components that decay to zero as the independent variable, usually \( x \), grows large. They are especially relevant in the study of dynamic systems, representing short-lived effects that vanish over time.

In our differential equation solution, the term \( Ce^{-2x} \) is identified as a transient term.

• As \( x \to \infty \), \( e^{-2x} \to 0 \), making the contribution of this term decrease steadily.
• Thus, it affects the solution temporarily and becomes negligible as \( x \) increases.

Transient terms are important to identify because understanding them helps anticipate how solutions stabilize in real-world applications.

They often represent initial conditions or transient behaviors occurring before a system settles into its long-term steady state."