Question

Find solutions \(y_{1}, y_{2}, \ldots, y_{n}\) of the equation \(y^{(n)}=0\) that satisfy the initial conditions $$ y_{i}^{(j)}\left(x_{0}\right)=\left\\{\begin{array}{ll} 0, & j \neq i-1, \\ 1, & j=i-1, \end{array} \quad 1 \leq i \leq n\right. $$

Short answer

Based on the step-by-step solution above, the final answer for the solutions \(y_1, y_2, \ldots, y_n\) of the given homogeneous linear differential equation with the provided initial conditions is: $$ y_1(x)=1, \\ y_2(x)=x, \\ y_3(x)=\frac{1}{2!x_0}x^2, \\ \cdots, \\ y_n(x)=\frac{1}{(n-1)!x_0^{n-1}}x^{n-1} $$

Step-by-step solution

Find the general solution of the homogeneous linear differential equation

The given equation is: $$ y^{(n)}(x) = 0 $$ To find the general solution, we first consider the characteristic polynomial, which is given by: $$ p(r)=r^n $$ Now, find the roots of this polynomial: $$ r^n = 0 \Rightarrow r=0 $$ It has \(n\) equal roots, so the general solution of the given equation is: $$ y(x)=C_{1}+C_{2}x+C_{3}x^2+\cdots+C_{n}x^{n-1} $$ where \(C_1, C_2, \ldots, C_n\) are arbitrary constants.

Apply the given initial conditions to find each constant

We must satisfy the initial conditions: $$ y_i^{(j)}\left(x_0\right)=\begin{cases} 0, & j \neq i-1,\\ 1, & j=i-1, \end{cases} \quad \text{for } 1 \leq i \leq n $$ For each solution \(y_i\), we differentiate the general solution \(i-1\) times to get: $$ y_i^{(i-1)}(x)=C_i (i-1)! x^{i-1} $$ Now, apply the initial condition \(y_i^{(i-1)}(x_0)=1\) and solve for \(C_i\): $$ C_i (i-1)! x_0^{i-1} = 1 \Rightarrow C_i = \frac{1}{(i-1)! x_0^{i-1}} $$ So the solution \(y_i\) is: $$ y_i(x) = \frac{1}{(i-1)! x_0^{i-1}} x^{i-1} $$

Write the final solutions

We have now found the solutions \(y_1, y_2, \ldots, y_n\) that satisfy the given initial conditions: $$ y_1(x)=1, \\ y_2(x)=x, \\ y_3(x)=\frac{1}{2!x_0}x^2, \\ \cdots, \\ y_n(x)=\frac{1}{(n-1)!x_0^{n-1}}x^{n-1} $$

Key concepts

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that meets specific conditions, usually given at a certain point. In the context of homogeneous differential equations, the solution must satisfy these initial conditions to be valid. For instance, when dealing with a differential equation like \[ y^{(n)} = 0 \]we often have conditions such as \[ y_i^{(j)}(x_0) = \begin{cases} 0, & j eq i-1, \ 1, & j = i-1, \end{cases} \]which means for each solution \(y_i\), its \((i-1)\)-th derivative at the point \(x_0\) is required to equal 1, while all other derivatives are zero.
These constraints are critical because they determine the form of the solution that matches the initial physical or conceptual system from which the differential equation originated. To solve an initial value problem, we often first solve the general form of the differential equation and then apply the initial conditions to find specific solutions.

Higher Order Derivatives

Higher order derivatives refer to derivatives taken multiple times. In problems involving differential equations, such as \(y^{(n)}(x) = 0\), repeated differentiation helps in expressing relationships among \(x\), \(y\), and their derivatives. Each order of derivative indicates a different aspect of the function's behavior, such as velocity, acceleration, etc., depending on the physical scenario.

In this context, solving a higher-order differential equation involves understanding its derivatives up to order \(n\). The important part of solving \(y^{(n)}(x)=0\) involves finding each of these functions \(y_i(x)\) that meets the initial conditions provided. Higher order derivatives often lead to complex relationships in real-world scenarios, but the principle remains the same: higher order derivatives seek answers to not just the function's present state but its rate, acceleration, and higher levels of change.

Characteristic Polynomial

The characteristic polynomial plays a central role when dealing with linear differential equations. It is often used to determine solutions to homogeneous differential equations. In the case of such an equation \(y^{(n)}(x) = 0\), the associated characteristic polynomial is \(p(r)=r^n\).

This polynomial effectively provides us with roots that help construct the solution to the differential equation. In our problem, all roots are \(r = 0\), indicating that the same operation is repeatedly applied, influencing how we derive the overall solution. Understanding and forming characteristic polynomials helps transform the differential equation solutions into an equation involving only algebraic roots, simplifying the solving process.

This concept simplifies the complex differential equations into more manageable algebraic problems, allowing us to find solutions efficiently through polynomial roots.

General Solution of Differential Equation

Solving a differential equation typically begins with finding its general solution. The general solution encompasses all potential solutions, defined by including constants that can be adjusted to meet specific conditions like initial values. For instance, the general solution for our homogeneous equation \(y^{(n)}(x)=0\) takes the form \(y(x)=C_{1}+C_{2}x+C_{3}x^2+ \cdots + C_{n}x^{n-1}\). This solution includes arbitrary constants \(C_1, C_2, \ldots, C_n\), which allow for adaptation based on initial conditions.

Once we have the general solution, applying the given initial conditions helps determine the specific values of these constants. This results in a unique solution that not only satisfies the differential equation but also matches the initial conditions accurately. Through this process, the general solution transforms into particular solutions, helping us understand how initial conditions shape the behavior of solutions of differential equations.