Problem 9 - Calculus and its applications

Question

Show that \(y=e^{x}+3 x e^{x}\) is a solution of $$ y^{\prime \prime}-2 y^{\prime}+y=0 $$

Step-by-Step Solution

Answer

The function \(y = e^x + 3xe^x\) satisfies the differential equation \(y'' - 2y' + y = 0\).

Step 1: Differentiate Once

First, find the first derivative of the function. Given the function \(y = e^x + 3xe^x\), apply the product rule to the \(3xe^x\) term. The derivative is:\[ y' = \frac{d}{dx}(e^x) + \frac{d}{dx}(3xe^x) = e^x + (3x)e^x + 3e^x = e^x + 3xe^x + 3e^x.\]Thus, we have:\[ y' = e^x + 3xe^x + 3e^x = e^x + 3e^x + 3xe^x. \]

Step 2: Differentiate Again

Next, find the second derivative. Differentiate the first derivative \(y' = e^x + 3e^x + 3xe^x\). Apply product rule again:\[ y'' = \frac{d}{dx}(e^x) + \frac{d}{dx}(3e^x) + \frac{d}{dx}(3xe^x) = e^x + 3e^x + (3e^x + 3xe^x) = e^x + 3e^x + 3e^x + 3xe^x.\]Thus, combining like terms:\[ y'' = e^x + 3e^x + 3e^x + 3xe^x = e^x + 6e^x + 3xe^x. \]

Step 3: Substitute into the Differential Equation

Substitute \(y''\), \(y'\), and \(y\) into the differential equation \(y'' - 2y' + y = 0\):1. Start with \(y'' = e^x + 6e^x + 3xe^x\).2. Substitute for \( -2y' = -2(e^x + 3e^x + 3xe^x) = -2e^x - 6e^x - 6xe^x\).3. Substitute \(y = e^x + 3xe^x\).Combine these:\[y'' - 2y' + y = (e^x + 6e^x + 3xe^x) - (2e^x + 6e^x + 6xe^x) + (e^x + 3xe^x).\]

Step 4: Simplify the Expression

Now, simplify the expression:\[y'' - 2y' + y = (e^x + 6e^x + 3xe^x) - (2e^x + 6e^x + 6xe^x) + (e^x + 3xe^x)\]Group and simplify terms:\[e^x + 6e^x + 3xe^x - 2e^x - 6e^x - 6xe^x + e^x + 3xe^x.\]Simplifying each component results in:\[0 = 0.\]Thus, it simplifies to zero, confirming \(y = e^x + 3xe^x\) is a solution to the differential equation.

Key Concepts

Solution Verification

In the study of differential equations, verifying a solution involves checking whether a suggested function satisfies a given differential equation. This process is crucial because it ensures that the proposed function is indeed a legitimate solution. The verification typically involves several steps:

• Calculate required derivatives of the function.
• Substitute these derivatives, along with the function itself, into the original differential equation.
• Simplify the expression to see if both sides of the equation balance.

When both sides equal zero or any constant value that they need to match, it confirms that the solution is correct. Therefore, in this exercise, the verification process showed that substituting the solution function into the differential equation resulted in zero, proving the proposed function is accurate.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes as its input changes. When solving differential equations, derivatives are crucial because they form the cornerstone of these types of equations. When a function involves exponential expressions, like in our problem with terms such as \(y = e^x + 3x e^x\), you need to find both the first and second derivatives. Here’s how:

• First Derivative: Use standard differentiation techniques, such as direct differentiation of simple terms and the product rule for composite terms.
• Second Derivative: Differentiate the first derivative to get higher order derivatives needed for substitution into the differential equation.

In our example, calculating both the first and second derivatives correctly allowed us to verify the solution by substitution into the original differential equation.

Product Rule

When working with functions in calculus, the product rule is an essential technique for differentiating terms that are the product of two or more functions. The rule states that the derivative of a product of two functions, \(u(x)\) and \(v(x)\), is given by:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]In the given exercise, the function term \(3x e^x\) required the use of the product rule to differentiate it correctly. Here's how it works:

• Identify \(u(x) = 3x\) and \(v(x) = e^x\).
• Calculate the derivative of both functions: \(u'(x) = 3\) and \(v'(x) = e^x\).
• Apply the product rule: \(\frac{d}{dx}[3x e^x] = 3e^x + 3x e^x\).

This application of the product rule ensures that complex expressions are accurately differentiated, essential for solving differential equations.

Homogeneous Equations

A homogeneous differential equation is one where every term is a function of the unknown variable and its derivatives. These equations are notable because they often simplify the process of finding general solutions. Given their structure, homogeneous equations like our example \(y'' - 2y' + y = 0\) have specific characteristics:

• They typically involve polynomial or exponential functions.
• Simplification can often lead to expressions that balance out to zero, confirming the solution.
• Solutions generally involve exponential functions that closely relate to the structure of the differential equation.

Solving these equations typically involves finding derivatives, substituting back, and checking if terms cancel out to confirm the solution. In this exercise, verifying that the function solved the homogeneous equation was crucial.