Problem 5 - Advanced Engineering Mathematics

Question

In Problems 5 and 6, compute \(y^{\prime}\) and \(y^{\prime \prime}\) and then combine these derivatives with \(y\) as a linear second-order differential equation that is free of the symbols \(c_{1}\) and \(c_{2}\) and has the form \(F\left(y, y^{\prime}, y^{\prime \prime}\right)=0\). The symbols \(c_{1}\) and \(c_{2}\) represent constants. $$ y=c_{1} e^{x}+c_{2} x e^{x} $$

Step-by-Step Solution

Answer

The differential equation is \( y'' - 2y' + y = 0 \).

Step 1: Differentiate the Function for First Derivative

Start with the given function: \[ y = c_1 e^x + c_2 x e^x \] Take the derivative of \(y\) with respect to \(x\) to find \(y'\). Apply the product rule on \(c_2 x e^x\), where \( u = x \) and \( v = e^x \). The derivative is:\[ y' = c_1 e^x + c_2 (e^x + x e^x) \] Simplify to:\[ y' = c_1 e^x + c_2 e^x + c_2 x e^x \] Combine like terms:\[ y' = (c_1 + c_2) e^x + c_2 x e^x \]

Step 2: Differentiate the Function for Second Derivative

Differentiate \(y'\) for the second derivative \(y''\).Starting with:\[ y' = (c_1 + c_2) e^x + c_2 x e^x \]Again, apply the product rule on \(c_2 x e^x\), where \( u = x \) and \( v = e^x \). The second derivative is:\[ y'' = (c_1 + c_2) e^x + c_2(e^x + x e^x) \]Simplify:\[ y'' = (c_1 + c_2) e^x + c_2 e^x + c_2 x e^x \]Combine terms:\[ y'' = (c_1 + 2c_2) e^x + c_2 x e^x \]

Step 3: Formulate the Differential Equation

Combine \(y\), \(y'\), and \(y''\) into a single equation free of \(c_1\) and \(c_2\):\[ y'' - 2y' + y = 0 \] Substitute \[ y = c_1 e^x + c_2 x e^x \], \[ y' = (c_1 + c_2) e^x + c_2 x e^x \], and \[ y'' = (c_1 + 2c_2) e^x + c_2 x e^x \].Upon substitution and simplification:\[ ((c_1 + 2c_2) e^x + c_2 x e^x) - 2((c_1 + c_2) e^x + c_2 x e^x) + (c_1 e^x + c_2 x e^x) \]Simplify the expression within the brackets:\[ = (0)e^x + (0)x e^x \]This confirms there are no terms left, meeting the condition \( F(y, y', y'') = 0 \).

Key Concepts

Linear Second-Order Differential Equations

A linear second-order differential equation is a type of equation involving the derivatives of a function. It is of the form \( a(x)y'' + b(x)y' + c(x)y = g(x) \), where

• \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \).
• \( g(x) \) is the non-homogeneous term or is zero for a homogeneous equation.

In this topic, the equation discussed is homogeneous because it is set to zero: \( y'' - 2y' + y = 0 \). This implies that the solution is determined entirely by the initial conditions or constants, rather than any additional external input represented by a non-zero \( g(x) \).
The goal is to express the relation among \( y \), \( y' \), and \( y'' \) such that the equation balances to zero. By eliminating the constants \( c_1 \) and \( c_2 \), we achieve an equation in its purest differential form.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to a variable. In our problem:

• The function \( y = c_1 e^x + c_2 x e^x \) is given.
• The first derivative, \( y' \), signifies how \( y \) changes as \( x \) changes.
• The second derivative, \( y'' \), gives insight into how \( y' \), the rate of change, changes itself.

When you differentiate the given function while keeping constants in mind, you use rules like the constant multiple rule and the sum rule. The process leads us to important simplifications needed to resolve the original equation. This method is crucial for understanding how physical and mathematical systems evolve, making it pivotal in fields such as physics and engineering.

Product Rule

The product rule is essential when differentiating products of two functions, helping find derivatives correctly in situations like the equation \( y = c_1 e^x + c_2 x e^x \).
Here, the product rule states that if you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product is:

• \( (uv)' = u'v + uv' \)

In our initial function, \( u = x \) and \( v = e^x \), so

• \( u' = 1 \)
• \( v' = e^x \).

Applying the product rule ensures each term in the equation is correctly differentiated, leading to the appearance and simplification of terms like \( c_2 e^x + c_2 x e^x \) in \( y' \) and \( y'' \). This rule proves indispensable in handling more complex functions, revealing their rates of change.

Exponential Functions

Exponential functions are a type of mathematical function characterized by an exponent which is a variable. They are in the form \( f(x) = a \, b^x \) where \( b \) is a constant base. In this context, the exponential function \( e^x \) is used.
Key characteristics of exponential functions are:

• Their derivative is proportional to the function itself, i.e., \( \frac{d}{dx} e^x = e^x \).
• Exponential growth is persistent and rapid due to the nature of its differentiation.

In mathematics, exponential functions are useful for modeling growth processes, compound interest, and more, as they describe systems that increase rapidly. Here, \( e^x \)'s property of having the same derivative is particularly helpful, ensuring that differentiations don't complicate to require successive steps. Understanding exponential functions is vital for mastering calculus and solving many real-world problems.