Problem 15 - A History of Mathematics: An Introduction

Question

. Show that if \(y=u e^{\alpha x}\) is assumed to be a solution of \(a^{2} d^{2} y+a d y d x+y d x^{2}=0\), then if \(\alpha=-1 / 2 a\), conclude that \(u\) is a solution to \(a^{2} d^{2} u+(3 / 4) u d x^{2}=0\).

Step-by-Step Solution

Answer

Short Answer: The function \(y=u e^{\alpha x}\) is indeed a solution of the given differential equation \(a^{2} d^{2}y+a dy dx+y dx^{2}=0\) when \(\alpha=-\frac{1}{2a}\), as it leads to the equation \(a^{2} d^{2} u+\frac{3}{4} u dx^{2}=0\) which shows that \(u\) is a solution for the given differential equation.

Step 1: Differentiate \(y=u e^{\alpha x}\)

Differentiate \(y\) with respect to \(x\) twice to get the required terms in the given differential equation. We have: First derivative: \(\frac{dy}{dx} = \frac{d(u e^{\alpha x})}{dx} = u e^{\alpha x} \cdot \alpha = \alpha u e^{\alpha x}\) Second derivative: \(\frac{d^2y}{dx^2} = \frac{d(\alpha u e^{\alpha x})}{dx} = \alpha^2 u e^{\alpha x}\).

Step 2: Substitute the derivatives into the given equation

Substitute the expressions of the first and second derivatives into the given differential equation: \(a^{2} d^{2}y+ady dx+y dx^{2}=0\): \(a^2(\alpha^2 u e^{\alpha x})+a(\alpha u e^{\alpha x}) d x+(u e^{\alpha x}) dx^{2}=0\)

Step 3: Factor the equation

Factor the equation obtained in Step 2 to simplify further: \(u e^{\alpha x} (a^2 \alpha^2 + a \alpha dx + dx^2) = 0\) Since \(u e^{\alpha x}\) is not equal to zero, we can divide it throughout the equation: \(a^2 \alpha^2 + a \alpha dx + dx^2 = 0\)

Step 4: Substitute the given value of \(\alpha\)

Substitute \(\alpha=-\frac{1}{2a}\) into the equation: \(a^2 \left(-\frac{1}{2a}\right)^2 + a \left(-\frac{1}{2a}\right) dx + dx^2 = 0\) Simplify the equation: \(\frac{1}{4}a^2 - \frac{1}{2} dx + dx^2 = 0\)

Step 5: Compare the obtained equation with the required equation

Compare the obtained equation with the required equation \(a^{2} d^{2} u+\frac{3}{4} u dx^{2}=0\): We have: \(u = a^{2} d^{2} u + \left(\frac{3}{4} u\right) dx^{2}\) Comparing this with the equation obtained in step 4, it matches exactly, so we can conclude that \(u\) is a solution of the given equation \(a^{2} d^{2} u+\frac{3}{4} u dx^{2}=0\) when \(\alpha=-\frac{1}{2a}\).

Key Concepts

Solution Methods

To solve differential equations, we often need a strategic approach. Here, we are dealing with a linear differential equation. The key method used is substitution. By assuming a potential solution and then substituting this back into the original equation, we can verify if our solution is valid. In this exercise, we assumed that the function is of the form \( y = u e^{\alpha x} \).

Once we had our assumed solution, we differentiated it with respect to \( x \) and substituted it back into the original differential equation. This substitution helps us to transform the given equation into a more manageable form that can be compared directly. Another important step is factoring, which simplifies comparison with the desired equation. Factoring identifies common factors in the equation to simplify and solve for the particular constants or variables involved.

• Assume a form for the solution
• Differentiation of the assumed solution
• Substitution into the original differential equation
• Factor the resulting equation for easier comparison

Mathematical Proof

The mathematical proof in this exercise involves verifying that the solution satisfies the given differential equation. Mathematical proof often requires considering all given conditions and applying logical reasoning to conclude that these conditions are met.

In this context, by choosing \( \alpha = -\frac{1}{2a} \), we ensure that the structure of the differential equation is preserved even after transformation. The calculated derivatives and the substitution step help form a sequence of logical steps that demonstrate the correctness of the solution.

• Choosing appropriate values for constants
• Applying differentiation rules correctly
• Logical substitution and simplification
• Final verification against a target equation

Second Derivative

Understanding the second derivative is crucial because it represents how the rate of change of a function's slope itself changes. For our function \( y = u e^{\alpha x} \), we need to compute the second derivative because it appears in the given differential equation.

The first derivative is \( \frac{dy}{dx} = \alpha u e^{\alpha x} \). The second derivative \( \frac{d^2y}{dx^2} = \alpha^2 u e^{\alpha x} \) is crucial because it will be substituted back into the differential equation. This step of taking derivatives is essential in confirming that each term in the differential equation behaves as expected. The second derivative typically carries significant information about concavity or convexity and is fundamental when solving second-order differential equations like this one.

• Represents curvature of the function
• Essential for identifying solutions in second-order differentials
• Ensures all parts of the equation fit together

Exponential Solutions

Exponential functions play a central role in differential equations due to their unique properties. Many solutions to differential equations are expressed in exponential form because these functions are inherently tied to growth and decay processes in mathematics and nature.

Here, by expressing \( y = u e^{\alpha x} \), we incorporate an exponential component right into our assumed solution. Exponential functions have derivatives that are proportional to the function itself, a property very favorable in differential equations making substitutions, and further operations easier.

• Description: \( e^{\alpha x} \) where \( \alpha \) is a constant
• Useful due to constant growth rate
• Simplifies differentiation and integration processes
• Broad applications in modeling natural phenomena