Problem 36 - Advanced Engineering Mathematics

Question

Use the concept that \(y=c,-\infty

Step-by-Step Solution

Answer

Yes, \(y = \frac{5}{3}\) is a constant solution.

Step 1: Understand the Problem

We have a second-order differential equation: \( y'' + 4y' + 6y = 10 \). The task is to determine if this equation has constant solutions by checking if \( y = c \) (a constant) satisfies this equation.

Step 2: Differentiate the Constant Function

For the function \( y = c \) to be a solution, its first derivative \( y' \) should be 0. This implies \( y'' \), the second derivative, must also be 0, since the derivative of a constant is zero.

Step 3: Substitute Constant into the Differential Equation

With \( y = c \), \( y' = 0 \), and \( y'' = 0 \), substitute into \( y'' + 4y' + 6y = 10 \) to see if this equation holds:\[ 0 + 4(0) + 6(c) = 10 \]

Step 4: Solve for the Constant c

The equation from Step 3 becomes:\[ 6c = 10 \]To solve for \( c \), divide both sides by 6:\[ c = \frac{10}{6} = \frac{5}{3} \]

Step 5: Check the Solution

Since the differential equation holds true with \( y = \frac{5}{3} \) when its derivatives are zero, \( y = \frac{5}{3} \) is a constant solution to the original differential equation.

Key Concepts

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a function. It typically looks like this: \( y'' + 4y' + 6y = 10 \).
The highest derivative, \( y'' \), is what defines it as second-order.
These equations often model dynamic systems where acceleration or curvature comes into play, such as in mechanics or population dynamics.
Key aspects include:

• Understanding both the first and second derivatives' effects on the solution.
• Recognizing that these equations can have multiple types of solutions — linear, constant, or oscillatory.

By mastering second-order differential equations, you'll be able to solve complex real-world problems.

Constant Solutions

Constant solutions in differential equations are solutions where the function \( y \) does not change with respect to the variable \( x \).
This means \( y \) remains constant, represented as \( y = c \).
When a solution is constant, its derivatives also equal zero.
This is because the rate of change of a constant value is zero.
To find constant solutions, you essentially test whether substituting \( y = c \) in the equation holds true when all derivatives are zero.

• Constant solutions are critical when you need stability in a system.
• They often represent equilibrium states in physical phenomena.

Differential Equation Solving

Solving a differential equation means finding all possible functions \( y(x) \) that satisfy the equation.
For our second-order differential equation, \( y'' + 4y' + 6y = 10 \), we explore whether specific forms of \( y(x) \) solve this equation.
In the case of constant solutions, we check if a number substituted for \( y \) works.

• Consider the nature of your function—constant, linear, or exponential.
• Simplify using substitutions where derivatives are known.

Improvements in your solving method come with practice and familiarity with similar forms of differential equations.

First Derivative

The first derivative, \( y' \), of a function reflects its rate of change.
For a constant function like \( y = c \), this derivative will always be zero.
Zero indicates no change—important for determining constant solutions.

• Measures immediate change rate of \( y \).
• Essential for understanding and solving many types of differential equations.

Recognizing the impact of the first derivative will help assess the dynamic behavior of functions.

Second Derivative

The second derivative, \( y'' \), describes the rate of change of the rate of change, or the acceleration of the function.
In our example, being zero along with the first derivative suggests a stable and unchanging solution.
When both derivatives are zero, this confirms that the function is constant.

• Indicates curvature or acceleration of a graph.
• Can reveal stability and oscillatory nature of systems.

Understanding second derivatives will deepen your insight into the behavior of functions beyond simple linear change.