Problem 26 - Differential Equations and Linear Algebra

Question

Determine two linearly independent solutions to the given differential equation on \((0, \infty)\) $$4 x^{2} y^{\prime \prime}+4 x(1+2 x) y^{\prime}+(4 x-1) y=0$$

Step-by-Step Solution

Answer

Two linearly independent solutions to the given differential equation on \((0, \infty)\) are \(y_1(x) = 1\) and \(y_2(x) = x\).

Step 1: Rewrite the given equation as a Cauchy-Euler equation

To rewrite the given equation as a Cauchy-Euler equation, divide both sides by \(4x^2\): \[ \begin{aligned} \frac{4 x^{2} y^{\prime \prime}}{4x^2}+\frac{4 x(1+2 x) y^{\prime}}{4x^2}+\frac{(4 x-1) y}{4x^2}=0 \Rightarrow y^{\prime\prime}+\frac{1+2x}{x}y'+\frac{4x-1}{4x^2}y=0 \end{aligned} \]

Step 2: Substitute \(y=x^m\) and its derivatives into the equation

To find the solutions to the Cauchy-Euler equation, consider a solution of the form \(y=x^m\), where \(m\) is a constant to be determined. Find the first and second derivatives of \(y\): \[y'=mx^{m-1}\] \[y''=m(m-1)x^{m-2}\] Substitute these values into the Cauchy-Euler equation obtained in Step 1: \[ \begin{aligned} m(m-1)x^{m-2}+\frac{1+2x}{x}mx^{m-1}+\frac{4x-1}{4x^2}x^m=0 \end{aligned} \]

Step 3: Solve for the characteristic equation

In order to remove the dependence on \(x\), we can multiply both sides of the equation by \(x^{2-m}\): \[ \begin{aligned} m(m-1) x^2+ (1+2x) m x + (4x - 1) &= 0 \end{aligned} \] Since this equation must be true for any \(x\) in the interval \((0,\infty)\), the coefficients must be equal to zero. Hence, \[ \begin{aligned} m(m-1)&=0 \end{aligned} \] which results in the characteristic equation: \(m^2 - m = 0\).

Step 4: Solve for \(m\) and obtain the linearly independent solutions

Solve the characteristic equation for \(m\) to determine the two linearly independent solutions: \[ \begin{aligned} m^2 - m &= 0 \\ m(m-1) &= 0 \\ m = 0,& m = 1 \end{aligned} \] The two linearly independent solutions are obtained by replacing \(m\) with the obtained values in the general form \(y=x^m\): For \(m=0\): \(y_1 = x^0 = 1\) For \(m=1\): \(y_2 = x^1 = x\) Thus, the two linearly independent solutions to the given differential equation on \((0, \infty)\) are \[y_1(x) = 1\] \[y_2(x) = x\]

Key Concepts

Linearly Independent Solutions

To determine the linearly independent solutions of a differential equation, we are looking for solutions that provide us with a complete basis for the solution space.
This is similar to having different ways to reach the same destination without overlap. If two solutions are linearly independent, one cannot be expressed as a multiple of the other.
**What does it mean for solutions to be independent?**

• If you have two solutions, say \( y_1(x) \) and \( y_2(x) \), they are linearly independent if a constant \( c \) cannot satisfy the equation \( c \, y_1(x) = y_2(x) \) for any non-zero value of \( c \).
• If solutions \( y_1 \) and \( y_2 \) can be linearly combined to create any possible solution to the differential equation, they form a basis of the solution space.

For the given exercise, the solutions \( y_1 = 1 \) and \( y_2 = x \) are linearly independent. This shows how, even within the framework of a seemingly tough equation, simple functions like these can solve the problem.

Characteristic Equation

The characteristic equation is a tool that helps us find the exponents which satisfy the differential equation. It emerges from substituting a proposed solution into the differential equation, greatly simplifying the problem of solving analytically cumbersome equations.

**How do we derive the characteristic equation?**

• For a Cauchy-Euler equation, we propose a solution of the form \( y = x^m \), where \( m \) is an unknown constant.
• Substitute this assumed solution into the differential equation; you'll get a polynomial equation in terms of \( m \).
• The roots of this polynomial are crucial—they provide the exponents for the linearly independent solutions.

In our exercise, the characteristic equation \( m^2 - m = 0 \) was derived by simplifying and removing the variable \( x \). Solving this provided the values \( m = 0 \) and \( m = 1 \), leading directly to our solutions \( y_1 = 1 \) and \( y_2 = x \). By finding \( m \), we step toward understanding the structure of the differential equation's solutions.

Differential Equations

Differential equations are foundational in mathematics, modeling many physical systems and natural phenomena. They describe how a certain quantity changes with respect to another, like time or space.

**Types and Solving Differential Equations**

• Differential equations can be ordinary (ODE) or partial (PDE), based on how many variables they depend on.
• ODEs, like the one in this exercise, involve functions of a single variable and their derivatives. Common techniques for solving these include finding characteristic equations or using special functions.

**Why are Cauchy-Euler Equations Special?**

• Cauchy-Euler equations are a particular type of ODE where coefficients are powers of the independent variable. They often transform into simpler equations using methods like substitution.
• They can be daunting at first but become more approachable with techniques like assuming polynomial solutions.

Understanding differential equations gives us the power to predict patterns and changes, tying together mathematical theory with real-world applications. This exercise specifically illustrates how structured approaches, like using a characteristic equation, provide direct routes to solutions.