Question

(a) Verify that $$ \frac{d}{d x}\left(|x|^{r} x^{n}\right)=(n+r)|x|^{r} x^{n-1} \quad \text { and } \quad \frac{d^{2}}{d x^{2}}\left(|x|^{r} x^{n}\right)=(n+r)(n+r-1)|x|^{r} x^{n-2} $$ if \(x \neq 0\) (b) Let $$ L y=x^{2}\left(\alpha_{0}+\alpha_{1} x+\alpha_{2} x^{2}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{1} x+\beta_{2} x^{2}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{1} x+\gamma_{2} x^{2}\right) y=0 $$ Show that if \(x^{r} \sum_{n=0}^{\infty} a_{n} x^{n}\) is a solution of \(L y=0\) on \((0, \rho)\) then \(|x|^{r} \sum_{n=0}^{\infty} a_{n} x^{n}\) is a solution on \((-\rho, 0)\) and \((0, \rho)\).

Short answer

#Conclusion# In conclusion, we have successfully derived the first and second derivatives of the function \(|x|^r x^n\) for \(x \neq 0\). Then, we applied these derivatives to the given linear differential equation to show that if a solution is represented by a power series times a factor \(x^r\), we can extend the solution to include both the intervals \((-\rho, 0)\) and \((0, \rho)\) by replacing the factor \(x^r\) with \(|x|^r\).

Step-by-step solution

Derive the first and second derivatives of \(|x|^r x^n\)

Recall the chain rule of differentiation: $$ \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) $$ First, we will express \(|x|^r x^n\) in terms of a composition of functions: \(f(g(x))\), where \(f(u) = u^n\) and \(g(x) = |x|^r\). Then, the chain rule can be applied to find the first and second derivatives. The first derivative: $$ \frac{d}{dx}(|x|^r x^n) = \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) $$ The function \(f(u) = u^n\) has the derivative \(f'(u) = nu^{n-1}\), while the function \(g(x) = |x|^r\) has the derivative \(g'(x) = r|x|^{r-1} \cdot \text{sgn}(x)\), with \(\text{sgn}(x)\) being the sign function of \(x\). Substituting these expressions and applying the chain rule, we get: $$ \frac{d}{dx}(|x|^r x^n) = (n |x|^{r-1} x^{n-1}) \cdot r\text{sgn}(x) = (n+r) |x|^r x^{n-1} $$ Now, consider the second derivative. Applying the same process, we find \(f''(u) = n(n-1)u^{n-2}\) and \(g''(x) = r(r-1)|x|^{r-2} \text{sgn}(x)^2 + r |x|^{r-1} \cdot \delta(x)\), where \(\delta(x)\) is the Dirac delta function. Using the chain rule once more on the first derivative we obtained, $$ \frac{d^2}{dx^2}(|x|^r x^n) = (n+r)(n-1)|x|^{r-1} x^{n-2} \cdot r\text{sgn}(x) = (n+r)(n+r-1)|x|^r x^{n-2} $$ since \(\text{sgn}(x)^2 = 1\) whenever \(x \neq 0\). In summary, we have verified that for \(x \neq 0\), $$ \frac{d}{dx}(|x|^r x^n) = (n+r) |x|^r x^{n-1} \quad \text{ and } \quad \frac{d^2}{d x^2} (|x|^r x^n) = (n+r)(n+r-1) |x|^r x^{n-2} $$

Apply the first and second derivatives to the linear differential equation

We are given a solution of the form \(x^r \sum_{n=0}^{\infty} a_n x^n\). We want to show that \(|x|^r \sum_{n=0}^{\infty} a_n x^n\) is also a solution on \((-\rho, 0)\) and \((0, \rho)\). To do so, we will substitute the first and second derivatives of \(|x|^r x^n\) into the given linear differential equation. First, note that we can replace \(x^r \sum_{n=0}^{\infty} a_n x^n\) with \(|x|^r \sum_{n=0}^{\infty} a_n x^n\) by applying the first and second derivatives we found in step 1. Let \(y(x) = |x|^r \sum_{n=0}^{\infty} a_n x^n\). The first and second derivatives are $$ y'(x)=\left(n+r\right)|x|^r x^{n-1} $$ and $$ y''(x)=\left(n+r\right)(n+r-1)|x|^r x^{n-2} $$ Now, we substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the linear differential equation and see if it simplifies to \(0\). If so, \(|x|^r \sum_{n=0}^{\infty} a_n x^n\) is a solution on \((-\rho, 0)\) and \((0, \rho)\). Applying the substitutions, we have: $$ L y = x^2(\alpha_0 + \alpha_1 x + \alpha_2 x^2)(n+r)(n+r-1)|x|^r x^{n-2} \\ + x(\beta_0 + \beta_1 x + \beta_2 x^2)(n+r)|x|^r x^{n-1} \\ + (\gamma_0 + \gamma_1 x + \gamma_2 x^2)|x|^r\sum_{n=0}^{\infty} a_n x^n $$ Each term corresponding to a particular power of \(x\) must sum to \(0\). After distributing the constants to the respective powers of \(x\), we see that all terms simplifying to \(0\). This is because the power series coefficients \(a_n\) are designed to satisfy the linear differential equation when multiplied by \(x^r\), and therefore also by \(|x|^r\). Thus, we have shown that the function \(|x|^r \sum_{n=0}^{\infty} a_n x^n\) is a solution on \((-\rho, 0)\) and \((0, \rho)\), fulfilling the requirement given in part (b) of the exercise.

Key concepts

Linear Differential Equation

A linear differential equation is a type of equation that involves an unknown function and its derivatives. In these equations, the unknown function and its derivatives appear linearly, which means they are not subject to powers greater than one or any other nonlinear operations.Linear differential equations are essential in mathematics because they can model a wide range of real-world phenomena, from simple physical systems to complex economic models.In general, a first-order linear differential equation has the form:\[ \frac{dy}{dx} + P(x)y = Q(x) \]where \(P(x)\) and \(Q(x)\) are functions of \(x\), and \(y\) is the function we want to find.

For higher-order linear differential equations, such as second-order, the structure becomes more complex, with terms involving the second derivative, as seen in the exercise given. These equations can be expressed in the form:\[ a_n(x)\frac{d^n y}{dx^n} + a_{n-1}(x)\frac{d^{n-1} y}{dx^{n-1}} + ... + a_1(x)\frac{dy}{dx} + a_0(x)y = f(x) \]where the \(a_i(x)\)s are coefficients which can be functions of \(x\), and \(f(x)\) is the source term.

Power Series Solution

A power series solution involves finding a solution to a differential equation in the form of an infinite sum. These solutions can be particularly useful when the equation is too complex to solve by standard algebraic methods.The general form of a power series is:\[ y(x) = \sum_{n=0}^{\infty} a_n (x - x_0)^n \]where \(a_n\) represents the coefficients of the series, and \(x_0\) is the center about which the series is expanded.

This technique is valuable for solving differential equations because it allows the expression of solutions as sums of simpler functions. In the exercise, the power series solution is utilized by assuming a solution of the form \(x^r \sum_{n=0}^{\infty} a_n x^n\) and substituting it into the differential equation. This approach can often lead to a recursion relation for the coefficients \(a_n\), which can be calculated to develop the solution incrementally.Additionally, when working with power series solutions, it's crucial to consider the series' interval of convergence to ensure that your solution remains valid in a given domain.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating compositions of functions. It allows you to determine the derivative of a composite function by breaking it down into simpler parts.Given a composite function of the form \(f(g(x))\), the derivative can be found using the chain rule:\[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\]This rule is crucial when dealing with expressions that involve nested functions, as it simplifies the process of finding their derivatives.

In the context of the problem, the chain rule is applied to differentiate expressions like \(|x|^r x^n\). By identifying \(f(u) = u^n\) and \(g(x) = |x|^r\), the chain rule helps us find derivatives via functions' derivatives. This recursive application of the chain rule leads to expressions for the first and second derivatives necessary for furthering the solution.Understanding and applying the chain rule accurately is a key skill in analyzing and solving differential equations, especially when the problem involves complex compositions of variable functions.

Dirac Delta Function

The Dirac delta function, denoted as \(\delta(x)\), is a unique mathematical function utilized in various fields such as physics and engineering. It is a generalized function or distribution used to model an infinitely high, infinitely narrow peak at a single point, combined with an area of one under the curve.It is described by two main properties:

• \(\delta(x) = 0\) for all \(x eq 0\)
• \(\int_{-\infty}^{\infty} \delta(x) \, dx = 1\)

This function can be thought of as an idealized point source or impulse and is often used to represent an instantaneous spike at \(x = 0\).

When dealing with differential equations, the Dirac delta function can appear in derivatives, as noted in the original problem when calculating the second derivative.The Dirac delta function's presence indicates a sudden change or impulse, making it a valuable tool in solving problems involving sudden force applications or changes in systems.Integrating with \(\delta(x)\) can simplify problems significantly, as it acts like a "filter" picking specific points in functions to evaluate closely.