Problem 42 - A First Course in Differential Equations with Modeling Applications

Question

Use the substitution \(t=x-x_{0}\) to solve the given differential equation. $$(x-4)^{2} y^{\prime \prime}-5(x-4) y^{\prime}+9 y=0$$

Step-by-Step Solution

Answer

The solution is \( y(x) = C_1(x-4)^3 + C_2(x-4)^3\ln(x-4) \).

Step 1: Identify the Substitution

Recognize that the given substitution is \( t = x - 4 \). This means that \( x = t + 4 \). This is crucial as it shifts the variable from \( x \) to \( t \), simplifying the differential equation.

Step 2: Rewrite the Derivatives

Compute the derivatives with respect to \( t \). First, note that \( \frac{d}{dt} = \frac{d}{dx} \times \frac{dx}{dt} = \frac{d}{dx} \). Hence, \( \frac{dy}{dt} = y' \) and \( \frac{d^2y}{dt^2} = y'' \). The derivatives remain the same because \( t \) and \( x \) only differ by a constant.

Step 3: Substitute in the Original Equation

Substitute \( t = x - 4 \) into the differential equation to get \((t)^2 y^{\prime\prime} - 5(t) y^{\prime} + 9y = 0\).

Step 4: Solve the Differential Equation

Recognize the form \((t)^2 y^{\prime\prime} - 5(t) y^{\prime} + 9y = 0\) is a Cauchy-Euler equation, which can be solved by assuming a solution \( y = t^m \). Substitute \( y = t^m \) into the equation to derive the characteristic equation \( m(m-1) - 5m + 9 = 0 \).

Step 5: Simplify the Characteristic Equation

Solve the characteristic equation: \( m(m-1) - 5m + 9 = 0 \) simplifies to \( m^2 - 6m + 9 = 0 \).

Step 6: Find the Roots of the Characteristic Equation

Factor the quadratic equation \( m^2 - 6m + 9 = 0 \) to get \( (m-3)^2 = 0 \), which means \( m = 3 \) is a repeated root.

Step 7: Construct the General Solution

Since \( m = 3 \) is a repeated root, the general solution of the differential equation is \( y(t) = C_1t^3 + C_2t^3\ln(t) \).

Step 8: Substitute Back to Original Variable

Finally, substitute \( t = x - 4 \) back into the general solution to express \( y \) in terms of \( x \): \( y(x) = C_1(x-4)^3 + C_2(x-4)^3\ln(x-4) \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are essential in modeling various physical phenomena, including motion, electricity, and heat flow. Understanding differential equations can help predict how systems evolve over time. In this exercise, we're dealing with a specific type of equation, the **Cauchy-Euler Equation**, known for its variable coefficients. These types of equations often change forms when variables change, which is why substitution methods come into play to simplify them.
The given equation is a second-order differential equation because it involves the second derivative of the function. Solving these equations usually involves several techniques like separation of variables, integration, and in this case, substitution. The ultimate goal is to find a function or set of functions that satisfy the equation, often referred to as the general solution.

Substitution Method

In differential equations, the substitution method is a powerful tool for simplifying problems. By changing variables or substituting values, complex equations can become more manageable. In our exercise, we used the substitution \( t = x - 4 \).
This shift in variables transforms the problem into a simpler form by centering it around the new variable \( t \).
The substitution helps to reveal hidden symmetries or simplify variable powers, making the equation easier to solve.

• Transformation: By replacing \( x \) with \( t+4 \) in derivatives, our variable of interest focuses around \( t \), leaving a function like \( (t)^2 y^{\prime\prime} - 5(t) y^{\prime} + 9y = 0 \).
• Ease of solution: It turns variable coefficients into constants and directly allows us to apply the Cauchy-Euler solution strategies.

This method is crucial in converting the original differential equation into a form that benefits from the famous solution approach related to Cauchy-Euler equations.

Characteristic Equation

The characteristic equation is a pivotal concept when handling linear differential equations. It arises from assuming a potential solution form, often using an exponential or power function. For Cauchy-Euler equations like in our exercise, we assume a solution of the form \( y = t^m \).
Upon substitution into the simplified differential equation, we derive a polynomial known as the characteristic equation. Here, substituting \( y = t^m \) transformed the equation to form:
\[m(m-1) - 5m + 9 = 0\]

• This equation is derived from equating the coefficients of like powers of \( t \) from the original equation.
• It is a quadratic equation, in our case, \( m^2 - 6m + 9 = 0 \), whose solutions, \( m \), determine the structure of the general solution for the differential equation.

Solving this characteristic equation is critical as it dictates the type of roots we encounter, whether real or complex, distinct or repeated, which in turn shapes the general form of the solution.

Repeated Roots

When solving the characteristic equation, we can encounter repeated roots. These occur when the polynomial's discriminant is zero, yielding a single repeated root solution. In our problem, we discovered that \(m = 3\) is a repeated root because
\[(m-3)^2 = 0\]resulted from factoring the characteristic equation.

Repeated roots lead to alternative solution structures compared to distinct roots. For repeated roots in a Cauchy-Euler context:

• The basic solution form derived from one root would be \( C_1t^m \), which is \( C_1t^3 \) in our exercise.
• An additional term is necessary to handle the repetition: using the natural logarithm, leading to \( C_2t^3\ln(t) \).

This incorporation of a logarithmic term ensures the general solution accommodates the mathematical properties of differential equations involving repeated roots. Such adaptations are crucial for ensuring the solution accurately reflects the behavior imposed by the differential equation constraints. In your final answer, expressing the solution in terms of \( x \) involves substituting back, giving \( y(x) = C_1(x-4)^3 + C_2(x-4)^3\ln(x-4) \). It shows the impact of the initial substitution on the solution's structure.