Problem 34 - 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 \((0, \infty)\). $$x^{2} y^{\prime \prime}+3 x y^{\prime}-8 y=0, x > 0$$

Step-by-Step Solution

Answer

The two linearly independent solutions to the given differential equation are \(y_1(x) = x^{-4}\) and \(y_2(x) = x^2\). Therefore, the general solution on (0, ∞) is \(y(x) = c_1x^{-4} + c_2x^2\), where c_1 and c_2 are arbitrary constants.

Step 1: 1. Rewrite the differential equation for y(x) = x^r

Rewriting the given differential equation for y(x) = x^r, we have: \(y = x^r\) \(y' = rx^{r-1}\) \(y'' = r(r-1)x^{r-2}\)

Step 2: 2. Plug it into the given differential equation and simplify

Now, substituting these expressions into the given differential equation: \(x^2(r(r-1)x^{r-2}) + 3x(rx^{r-1}) - 8(x^r) = 0\) Simplify the equation: \(r(r-1)x^r + 3rx^r - 8x^r = 0\)

Step 3: 3. Determine the values of r that result in linearly independent solutions

Now, we will factor out x^r from the equation to find possible values of r: \((r(r-1) + 3r - 8)x^r = 0\) For the equation to hold true for all x > 0, the term inside parentheses must be equal to 0: \(r(r-1) + 3r - 8 = 0\) This is a quadratic equation in r. Let's solve it for r.

Step 4: 4. Solve the quadratic equation for r

To solve the quadratic equation, we first simplify it: \((r^2 - r + 3r - 8) = 0\) \((r^2 + 2r - 8) = 0\) Now, factor the equation: \((r+4)(r-2) = 0\) Thus, we have two possible values for r: 1. \(r_1 = -4\) 2. \(r_2 = 2\)

Step 5: 5. Write the linearly independent solutions and the general solution

Using our two values of r, we can now write the linearly independent solutions y_1(x) and y_2(x) for the given differential equation: 1. \(y_1(x) = x^{-4}\) 2. \(y_2(x) = x^2\) Finally, we can write the general solution for the given differential equation on the interval (0, ∞): \(y(x) = c_1x^{-4} + c_2x^2\) where c_1 and c_2 are arbitrary constants.

Key Concepts

Linearly Independent Solutions

When dealing with differential equations, it is often necessary to find solutions that are linearly independent. But what does "linearly independent" mean in this context? Simply put, two functions are linearly independent if there is no constant that can multiply one to make it the same as the other—except for the trivial case where the constant is zero. This property assures us that the solutions capture all the possible behaviors of the differential equation.

In our example, we calculated two such solutions:

• \(y_1(x) = x^{-4} \)
• \(y_2(x) = x^2 \)

These solutions do not conflict with each other in any way that would make one a mere constant multiple of the other. This is important for constructing what we call the general solution, as having linearly independent solutions allows us to cover a wider range of potential solutions.

General Solution

The general solution of a differential equation encapsulates all possible specific solutions you might encounter. By combining the linearly independent solutions we've found, and ensuring each has an arbitrary constant, we create the general form. In our case, this meant we could combine the solutions we found into the form

• \(y(x) = c_1 x^{-4} + c_2 x^2 \)

where \(c_1\) and \(c_2\) are constants that can take any value.

These constants provide flexibility, allowing this solution to adapt to boundary conditions or initial values specified in a problem. As you'll often find in differential equation work, being able to form the general solution is critical for understanding the complete set of behaviors described by the equation.

Quadratic Equation in Parameters

Sometimes solving differential equations involves solving a subsidiary equation, such as a quadratic equation that comes from plugging an assumed solution into the original equation. In our case, we plugged in functions of the form \(y(x) = x^r\), leading us to derive a quadratic in \(r\):

• \(r^2 + 2r - 8 = 0\)

Solving this quadratic helped us find the values for \(r\) — specifically \(-4\) and \(2\) — enabling us to identify our linearly independent solutions.

The process of solving the quadratic equation might involve factoring or using the quadratic formula, both useful skills for solving physics and engineering problems. In essence, find these roots to determine how your solutions to the differential are composed.

Second Order Differential Equation

The differential equation we grapled with belongs specifically to the second order, meaning it involves the second derivative of a function. The general form for such an equation might look like:

• \(a(x) y'' + b(x) y' + c(x) y = 0\)

In our example, \(a = x^2\), \(b = 3x\), and \(c = -8\). These types of differential equations are a bit more complicated than first-order ones, involving more complex methods like assuming a solution of a particular form and solving for coefficients.

Second order differential equations can describe a variety of phenomena from physics to finance. In this exercise, getting familiar with their structure and solution process will empower you to tackle more intricate models.