Problem 35 - Differential Equations and Linear Algebra

Question

Determine two linearly independent solutions to the given differential equation of the form \(y(x)=x^{r},\) and thereby determine the general solution to the differential equation on \((0, \infty)\). $$2 x^{2} y^{\prime \prime}+5 x y^{\prime}+y=0, x > 0$$

Step-by-Step Solution

Answer

The general solution to the given differential equation on the interval \((0, \infty)\) is: \[ y(x) = C_1x^{-\frac{1}{2}} + C_2x^{-1}, \] where \(C_1\) and \(C_2\) are arbitrary constants.

Step 1: Substituting the suggested form of the solution

Let's first substitute the suggested form of the solution, \(y(x) = x^{r}\), into the given differential equation. To do this, we need to find the first and second derivatives of \(y(x)\) as well \[ y'(x) = rx^{r-1}, \] \[ y''(x) = r(r-1)x^{r-2}. \] Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given equation: \[ 2x^2(r(r-1)x^{r-2}) + 5x(rx^{r-1}) + x^r = 0. \]

Step 2: Simplify and factor the equation

Next, let's simplify the equation and factor out the common term \(x^r\): \[ 2r(r-1)x^r + 5rx^r + x^r = x^r(2r^2 - 2r + 5r + 1) = x^r(2r^2 + 3r +1)= 0. \] Since \(x > 0\), the only way this equation can be true is if the expression in the parentheses is zero: \[ 2r^2 + 3r + 1 = 0. \]

Step 3: Solve the quadratic equation for r

Now, we need to solve the quadratic equation for \(r\): \[ 2r^2 + 3r + 1 = 0. \] This equation can be factored into \((2r+1)(r+1) = 0\). From this factorization, we can find the two solutions for \(r\): \[ r_1 = -\frac{1}{2}, \quad r_2 = -1. \]

Step 4: Find two linearly independent solutions

Using the values of \(r\), we can find two linearly independent solutions \(y_1(x)\) and \(y_2(x)\): \[ y_1(x) = x^{r_1} = x^{-\frac{1}{2}}, \quad y_2(x) = x^{r_2} = x^{-1}. \]

Step 5: Determine the general solution

To find the general solution to the given differential equation, we can use the superposition principle, which states that any linear combination of the independent solutions will also be a solution. Therefore, the general solution of the equation can be expressed as: \[ y(x) = C_1y_1(x) + C_2y_2(x) = C_1x^{-\frac{1}{2}} + C_2x^{-1}. \] Thus, the general solution to the given differential equation on the interval \((0, \infty)\) is: \[ y(x) = C_1x^{-\frac{1}{2}} + C_2x^{-1}, \] where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Linearly Independent Solutions

In the realm of differential equations, particularly for second-order types, finding solutions is crucial to understanding the behavior of the system described by the equation. A linearly independent solution means that two functions, in this context, do not simply differ by a constant multiple. Instead, they bring different behaviors or structures to the model. For our differential equation:

• Two solutions we found were: \(y_1(x) = x^{-\frac{1}{2}}\) and \(y_2(x) = x^{-1}\).
• These solutions fundamentally differ in their form and cannot be recreated by merely multiplying one by a constant to produce the other.

To test for linear independence, we often verify these solutions using the Wronskian determinant. If the Wronskian (denoted as \(W(f,g)\)) is not zero at any point, the functions \(f(x)\) and \(g(x)\) are linearly independent.

General Solution

The general solution of a second-order differential equation is a crucial concept because it aggregates all possible solutions. It enables us to describe any specific solution to an equation by modifying parameters. Considering the differential equation at hand, the general solution is expressed as:

• \(y(x) = C_1x^{-\frac{1}{2}} + C_2x^{-1}\).

This form showcases the beauty and flexibility imparted by the linearity of the equation, allowing us to generate countless specific solutions by altering the constants \(C_1\) and \(C_2\). These constants adapt to initial conditions, thus encompassing a complete family of solutions for the given differential context. In essence, the general solution is the most exhaustive depiction of solutions for the equation you are solving.

Superposition Principle

The superposition principle is a mathematical merriment in linear differential equations. It allows for combining solutions, showcasing their linear nature. A fundamental property of any linear system, the principle states:

• If \(y_1(x)\) and \(y_2(x)\) are solutions to a linear homogeneous differential equation, any linear combination \(C_1y_1(x) + C_2y_2(x)\) will still be a solution.

For our problem, we used this principle to construct the overall form of the general solution. By taking two linearly independent solutions, \(y_1(x) = x^{-\frac{1}{2}}\) and \(y_2(x) = x^{-1}\), and combining them with arbitrary constants \(C_1\) and \(C_2\), we ensured that the most general solution was captured, suited for a wide variety of initial conditions and scenarios.

Second-Order Differential Equations

Second-order differential equations hold significant applicability in modeling various physical systems like oscillations, waves, or electrostatics. These equations involve the second derivative of an unknown function, which often represents acceleration or curvature in real-world phenomena.

• Our specific equation \(2 x^2 y'' + 5 x y' + y = 0\) is an example of a linear homogeneous second-order differential equation.
• These equations characteristically lead us to find two independent solutions to form the general solution.

The methods to solve such equations routinely involve identifying a characteristic equation, which helped us determine the possible values of \(r\) for which \(y(x) = x^{r}\) becomes a viable solution. This understanding paves the way for analyzing system behavior under various input conditions and responses.