Question

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 8 x^{2}\left(2-x^{2}\right) y^{\prime \prime}+2 x\left(10-21 x^{2}\right) y^{\prime}-\left(2+35 x^{2}\right) y=0 $$

Short answer

Question: Find the general solution to the following second-order linear differential equation with variable coefficients: $$ 8x^2 (2 - x^2) y'' + 2x (10 - 21x^2) y' - (2 + 35x^2) y = 0 $$ Answer: The general solution to the given differential equation is: $$ y(x) = C_1 x^{\frac{1}{4}} + C_2 x^{-\frac{5}{2}} $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Identify the substitution and rewrite the equation

Suggested substitution to consider is \(y(x) = x^m\). This means: $$ y'(x) = m x^{m-1} \\ y''(x) = m(m-1) x^{m-2} $$ Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) with these expressions into the original equation: $$ 8x^2(2-x^2)m(m-1)x^{m-2}+2x(10-21x^2)m x^{m-1}-(2+35x^2)x^m=0 $$

Simplify the equation and find exponent m

Next, we should simplify the equation. In order to do that, we can factor out \(x^m\) from every term: $$ x^m[8(2-x^2)m(m-1)x^{-2}+2(10-21x^2)m x^{-1}-(2+35x^2)]=0 $$ For non-trivial solutions, we can now divide by \(x^m\): $$ 8(2-x^2)m(m-1)x^{-2}+2(10-21x^2)m x^{-1}-(2+35x^2)=0 $$ Now, to find the value of the exponent \(m\), we will assume that the expression inside the brackets is an identity in \(x\), meaning that it should hold true for all values of \(x\): $$ 8(2)m(m-1)+2(10)m-(2)=0 $$ Solve this equation for \(m\): $$ m= \frac{1}{4}, -\frac{5}{2} $$

Find the two independent solutions

Now that we have obtained two values for exponent \(m\), let us find the two linearly independent solutions corresponding to these values: Solution 1: For \(m = \frac{1}{4}\), $$ y_1(x) = x^{\frac{1}{4}} $$ Solution 2: For \(m = -\frac{5}{2}\), $$ y_2(x) = x^{-\frac{5}{2}} $$

Write down the general solution

Now that we have two independent solutions of our differential equation, we can write down the general solution to our problem on some interval \((0,\rho)\): $$ y(x) = C_1 x^{\frac{1}{4}} + C_2 x^{-\frac{5}{2}} $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

General Solution

Understanding the general solution of a differential equation is akin to grasping the 'big picture' of what the equation is expressing. In essence, the general solution captures all possible solutions to a differential equation.

Consider a scenario where we have myriad paths to reach a particular destination; similarly, a differential equation can have multiple solution paths or functions that satisfy it. However, these solutions are not arbitrary; they often include constants, like our variables C₁ and C₂, which can take on any value, allowing for infinite possibilities that cover all potential scenarios represented by the equation.

It's important to note that the general solution incorporates all particular (specific) solutions and also honors any initial conditions or boundaries provided, thus painting a complete solution landscape. In our case, the general solution has the form y(x) = C₁ x^{1/4} + C₂ x^{-5/2}, which represents an infinite set of solutions depending on the values of C₁ and C₂.

Substitution Method

When we tackle complex differential equations, the substitution method can be a beacon of simplification. It's akin to swapping out an intricate piece of a puzzle with a more manageable counterpart.

The essence of this technique is to reduce the complexity of the original equation by replacing a function or its derivatives with a simpler expression that retains the fundamental properties. By doing this, we can often transform a daunting equation into a more manageable form.

In the exercise at hand, we replaced the function y(x) with x^m and its derivatives accordingly. This approach let us simplify the problem and laid out a clear path forward to finding the powers of x that would satisfy the equation, ultimately leading to the general solution.

Ordinary Differential Equations

Dive into the realm of Ordinary Differential Equations (ODEs), and you're exploring the relationships between functions and their derivatives. These equations are the bedrock for describing dynamic systems, from the orbits of celestial bodies to the oscillations of a pendulum.

What sets ODEs apart is that they involve functions of a single variable and their derivatives. The order of an ODE corresponds to the highest derivative present; our focus is on second-order ODEs since we have a term with y''(x) as the highest derivative involved. Successful analysis of such equations allows predictions about system behavior and, in practical terms, enables us to engineer solutions and understand natural phenomena.

Linear Independence

In the symphony of mathematics, linear independence is like distinct musical notes that, when combined, produce a rich, full sound. Linear independence in the context of solutions to differential equations means that no solution is a mere echo or multiple of another; each solution is unique and contributes something new.

To claim that functions are linearly independent, particularly in the context of differential equations like the one highlighted, one must be unable to express any one of the functions as a combination of the others using multiplication by constants. This uniqueness is essential when constructing the general solution from a set of particular solutions, ensuring that the general solution covers all possible scenarios without redundancy.

For our example, the solutions resulting from different values of m are linearly independent. Therefore, the general solution y(x) = C₁ x^{1/4} + C₂ x^{-5/2} holds true.

Boundary Value Problems

Boundary value problems (BVPs) serve as the guardrails on the road to finding specific solutions to differential equations. They provide crucial constraints that ensure the solutions we find are precisely tailored to the physical or geometric conditions of the scenario in question.

These problems are distinguished by specifying values or relationships that the solution must satisfy at distinct points, commonly referred to as 'boundaries'. In practical situations, this could mean setting the temperature at the ends of a rod, the displacement of a string at its endpoints, or the electric potential on the surface of a conductor.

Applying BVPs turn the general solution of a differential equation into a particular solution that fulfills the stipulated boundaries. However, in our example, we're focused on finding the general solution which later can be tailored with given boundary conditions to yield a specific solution.