Question

Suppose that \(f\) is continuous on \([a, \infty)\), $$ y^{(n)}(x)=f(x), \quad t \geq a $$ and \(y(a)=y^{\prime}(a)=\cdots=y^{(n-1)}(a)=0 .\) (a) Integrate repeatedly to show that $$ y(x)=\int_{a}^{x} d t_{n} \int_{a}^{t_{n}} d t_{n-1} \cdots \int_{a}^{t_{3}} d t_{2} \int_{a}^{t_{2}} f\left(t_{1}\right) d t_{1} $$ (b) By successive reversals of orders of integration as in Example \(7.2 .11,\) deduce from (A) that $$ y(x)=\frac{1}{(n-1) !} \int_{a}^{x}(x-t)^{n-1} f(t) d t $$

Short answer

Question: Determine the function \(y(x)\) that satisfies the given conditions: \(y^{(n)}(x) = f(x)\), \(y(a) = 0\), and \(y'(a) = y''(a) = \cdots = y^{(n-1)}(a) = 0\), where \(f\) is a continuous function on \([a, \infty)\) and \(n\) is a positive integer. Answer: The function \(y(x)\) that satisfies the given conditions is $$ y(x) = \frac{1}{(n-1)!}\int_{a}^{x}(x-t)^{n-1} f(t) dt. $$

Step-by-step solution

Integrate the nth Order Differential Equation

Given, \(y^{(n)}(x) = f(x)\). To find \(y(x)\), we need to integrate \(f(x)\) repeatedly \(n\) times. Integrating the first time, we get: $$ y^{(n-1)}(x) = \int f(x) dx + C_1 $$ Note that the integration constant \(C_1\) will be eliminated in the next integration. #a:Integrate Repeatedly#

Repeated Integrations

Next, we integrate the \((n-1)\)th order differential equation \(n-1\) times. For each integration, we perform a change of variable, starting with \(t_n = x\), then \(t_{n-1}\) and so on. After integrating \(n\) times, we will arrive at the following expression: $$ y(x) = \int_{a}^{x} dt_n \int_{a}^{t_n} dt_{n-1} \cdots \int_{a}^{t_3} dt_2 \int_{a}^{t_2} f(t_1) dt_1 $$ #b:Change the Order of Integration#

Change the Order of Integration

Now we will change the order of integration as indicated in Example 7.2.11. Through successive reversals of order, we can rewrite the integral above as: $$ y(x) = \int_{a}^{x} dt_1 \int_{t_1}^{x} dt_2 \cdots \int_{t_{n-2}}^{x} dt_{n-1} \int_{t_{n-1}}^{x} f(x) dx $$ #b:Observe Similar Patterns#

Observe Similar Patterns and Simplify

We can observe that there is a pattern of integration limits. In each integral, the upper limit is \(x\), and the lower limit is the next variable: \(t_1\), \(t_2\), ..., \(t_{n-1}\). Thus, each integration factor results in a term of the form \((x-t)^p\), where \(p\) is the power of the term. After performing \(n-1\) of these integrations, the power of the term will be \((x-t)^{n-1}\). Also, note that there will be a factorial factor due to the multiple integrations. This is accumulated through each integration, and the final result includes the factor \((n-1)!\). Combining these observations, we can simplify the integral expression for \(y(x)\) as: $$ y(x) = \frac{1}{(n-1)!}\int_{a}^{x}(x-t)^{n-1} f(t) dt $$ This is the final solution for the given problem.

Key concepts

Continuous Functions

In mathematics, continuous functions play a crucial role, especially when dealing with differential equations. A function is continuous over an interval if there are no breaks, jumps, or points of discontinuity within that interval. Simply put, you can draw the function without lifting your pencil from the paper.
Continuous functions ensure smooth behavior of the equations and solutions they are involved in. For example, consider a function \( f(x) \) that is continuous on the interval \([a, \infty)\). This means that for any number \( x \) within that interval, \( f(x) \) does not abruptly change its value.

• Ensures consistent and predictable results
• Guarantees differentiability within the interval
• Important for solving differential equations as it helps in using integration techniques effectively

When dealing with differential equations, starting with a continuous function guarantees that integrations will proceed smoothly without surprises. This concept is foundational when integrating consecutive orders of derivatives, as each order builds upon the smooth nature of the previous ones.

Integration Techniques

Integration is a powerful mathematical tool used to find the function that describes the accumulative effect of a rate of change, represented by a differential equation. The ability to integrate a function repeatedly is especially useful in differential equations, where we start with a derivative and need to find the original function.
In this exercise, repeated integration was performed to solve a higher-order differential equation. The process can be broken down as follows:

• Begin integrating the highest order derivative down to a lower order.
• Apply successive integrations to iteratively reduce the order until reaching the original function.
• The technique requires careful application of limits and variables, often involving substitution methods to simplify each step.

It’s important to note that each integration may introduce a constant of integration, which is often determined by boundary conditions provided, such as \( y(a) = 0 \), to tailor the general solution to the specific problem.
Lastly, changing the order of integration can simplify the resulting integral significantly. This was utilized here to deduce the final expression by recognizing patterns and re-evaluating the integration limits.

Order of Integration

The order of integration pertains to the sequence in which integrations are performed when executing multiple integrations. It is particularly significant in cases where repeated integration is employed to solve differential equations, as demonstrated in this problem.
When integrating a function multiple times, the order you choose can affect the ease with which the task is accomplished. Initially, we integrated starting from the outermost variable and working our way inwards, which aligned with the initial format given in the exercise.

• Understanding and manipulating the order can significantly simplify calculations.
• Reversing the order, as seen here, was effective in simplifying the final result.
• It allows the restructuring of the problem into a more manageable form, often revealing patterns such as factorial terms or powers of variables.

In our specific example, by reversing the order of integration, we deduced that \( y(x) = \frac{1}{(n-1)!} \int_{a}^{x}(x-t)^{n-1} f(t) dt \) was a more efficient and straightforward form. Such strategic reorderings are essential in efficiently reaching the solution while ensuring the integrations are executed correctly.