Question

We've shown that if \(p\) and \(f\) are continuous on \((a, b)\) then every solution of $$ y^{\prime}+p(x) y=f(x) $$ on \((a, b)\) can be written as \(y=u y_{1},\) where \(y_{1}\) is a nontrivial solution of the complementary equation for \((\mathrm{A})\) and \(u^{\prime}=f / y_{1} .\) Now suppose \(f, f^{\prime}, \ldots, f^{(m)}\) and \(p, p^{\prime}, \ldots, p^{(m-1)}\) are continuous on \((a, b),\) where \(m\) is a positive integer, and define $$ \begin{array}{l} f_{0}=f \\ f_{j}=f_{j-1}^{\prime}+p f_{j-1}, \quad 1 \leq j \leq m \end{array} $$ Show that $$ u^{(j+1)}=\frac{f_{j}}{y_{1}}, \quad 0 \leq j \leq m $$

Short answer

Question: Prove that for the given first-order linear ordinary differential equation and conditions on the functions \(f\) and \(p\), the relationship \(u^{(j+1)}=\frac{f_{j}}{y_1}, \quad 0 \leq j \leq m\) holds. Answer: Using mathematical induction, we proved the relationship \(u^{(j+1)}=\frac{f_{j}}{y_1}, \quad 0 \leq j \leq m\) for given conditions on the functions \(f\) and \(p\). This proof involved recalling important definitions and notations, proving the relationship for the base case (\(j = 0\)), assuming the relationship holds for some \(j = k\), and then showing that it also holds for \(j = k+1\).

Step-by-step solution

Recall Important Definitions and Notations

Recall that for this given first-order linear ordinary differential equation (ODE), $$ y^{\prime}+p(x)y=f(x). $$ \(y_1\) is a nontrivial solution of the complementary equation, which means it is a solution of the homogeneous equation: $$ y^{\prime}+p(x)y=0. $$ And the integrating factor \(u\) satisfies \(u^{\prime}=\frac{f}{y_1}\). Now, we have a series of functions \(f_j\) with the following relationship: $$ \begin{array}{l} f_{0}=f \\\ f_{j}=f_{j-1}^{\prime}+p f_{j-1}, \quad 1 \leq j \leq m \end{array}. $$

Prove the Relationship for \(j = 0\)

First, let's prove that the relationship \(u^{(j+1)}=\frac{f_{j}}{y_{1}}, \quad 0 \leq j \leq m\) holds true for \(j = 0\). In this case, it is given that $$ u^{\prime} = \frac{f}{y_1} = \frac{f_0}{y_1}. $$ Thus the relationship holds for \(j = 0\).

Use Induction Hypothesis and Derivative Properties

Now, let's assume that the relationship holds for some \(j = k\), where \(0 \leq k < m\). That is, $$ u^{(k+1)} = \frac{f_{k}}{y_1}. $$ We will prove that the relationship also holds for \(j = k+1\). Recall that $$ f_{k+1} = f_{k}^{\prime} + p f_k. $$ We can differentiate both sides of our induction hypothesis with respect to \(x\): $$ (u^{(k+1)})^{\prime} = \frac{d}{dx}\left(\frac{f_k}{y_1}\right). $$

Derive the Relationship for \(j = k + 1\)

Using the quotient rule, we have $$ (u^{(k+1)})^{\prime}=\frac{f_{k}^{\prime}y_{1}-f_{k}y_{1}^{\prime}}{y_{1}^{2}}. $$ But \(y_1\) is a solution of the complementary equation, so \(y_{1}^{\prime} = -p y_1\). Substitute this into the above equation, we get $$ (u^{(k+1)})^{\prime}=\frac{f_{k}^{\prime}y_{1}+p f_{k}y_{1}}{y_{1}^{2}}=\frac{f_{k+1}}{y_1}. $$ This shows that \(u^{(k+2)}=\frac{f_{k+1}}{y_1}\), completing the induction step.

Conclusion

By the principle of mathematical induction, we have shown that the relationship $$ u^{(j+1)}=\frac{f_{j}}{y_1}, \quad 0 \leq j \leq m $$ holds for all \(0 \leq j \leq m\), given the conditions specified in the problem statement about the functions \(f\) and \(p\).

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation where the equation contains only the first derivative of the unknown function. It has the general form: \( y' + p(x)y = f(x) \). This means that it involves a function \( y \), its first derivative \( y' \), and functions \( p(x) \) and \( f(x) \) of the independent variable \( x \). These equations model a wide range of real-world phenomena, from population growth to electrical circuits.

To solve such an equation, you often find an integrating factor, which transforms the differential equation into one that can be solved more easily. This method involves multiplying the entire equation by a specific function that makes the left-hand side a perfect derivative, thus allowing integration to solve for \( y \).

Solving first-order linear differential equations can be a robust tool in mathematics because these equations appear frequently in both theoretical and practical applications.

Complementary Equation

In solving differential equations, particularly linear differential equations, the complementary equation plays a critical role. The complementary equation is derived from the original by setting the nonhomogeneous part \( f(x) \) to zero. For our equation:
\( y' + p(x)y = f(x) \), the complementary equation is: \( y' + p(x)y = 0 \).

This equation is also called the homogeneous equation. It reflects solutions arising solely from the influence of the system itself, without external force or input.

Finding the solution to the complementary equation (often denoted as \( y_1 \)) is crucial because it helps construct the general solution for the original differential equation. This is done by adding the particular solution (solving the whole equation with \( f(x) \)), \( y_p \), to the complementary solution, forming the total solution \( y = y_1 + y_p \).

The complementary equation therefore helps in understanding the natural behavior pattern of the system described by the differential equation.

Mathematical Induction

Mathematical induction is a proof technique used to establish a given statement for all natural numbers. It's a handy tool in many areas of mathematics, including solving differential equations. This approach involves two main steps:

• **Base Case**: Show that the statement holds for the initial value (typically \( j = 0 \)).


• **Induction Step**: Assume the statement holds for some integer \( k \). Then demonstrate it holds for \( k+1 \).

In our differential equation context, mathematical induction shows that a certain property or solution, hypothesized to work for \( j = 0 \), works for all subsequent indices \( j \). It's like a domino effect—if one falls (if the base case is true), they all continue falling (the induction step).

This method is particularly useful in proving sequences of equations, such as the relationship \( u^{(j+1)} = \frac{f_j}{y_1} \) for all \( 0 \leq j \leq m \). It's a systematic, reliable way to confirm the correctness of conceptual mathematical relationships over an infinite sequence.

Homogeneous Equation

A homogeneous equation is a specific kind of differential equation where the equation equals zero. In the realm of first-order linear differential equations, the homogeneous form is:
\( y' + p(x)y = 0 \).

This form arises when you set the function \( f(x) \) to zero, essentially removing any external forces affecting the system described by the differential equation.

Homogeneous equations are vital because they give insight into the inherent behavior of a system. Solving this equation provides a fundamental solution \( y_1 \) which is used to build the general solution for more complex, non-homogeneous equations. It represents the solution related only to the system's structural properties and internal conditions.

When solved, homogeneous equations often provide solutions in exponential form, such as \( y = Ce^{- ext{some function of } x} \), where \( C \) is a constant. This kind of solution tells how systems, like physical phenomena, can evolve over time without external influences affecting them.