Solve Problems by use of Fourier series. Assume in each case that the right- hand side is a periodic function whose values are stated for one period. $$y^{\prime \prime}+9 y=\left\\{\begin{array}{ll} x, & 0

To break down any periodic function into a sum of sines and cosines, we need to find its Fourier coefficients. These coefficients tell us how much of each sine and cosine wave is in the function.

The function we are dealing with is defined as follows for one period:

• For $$0 < x < 1$$, $$f(x) = x$$.
• For $$-1 < x < 0$$, $$f(x) = 0$$.

The Fourier coefficients can be found using integration. The formula for the constant term (which does not depend on $$x$$) is:
\[ a_0 = \frac{1}{2} \frac{1}{L} \bigg( \int_{-L}^{L} f(x) dx \bigg) \]
Here, $$L$$ is half the period. Our function is odd, which means that the integral over half the period is twice the integral over the positive half. So we calculate the integral from 0 to 1 and then multiply by 2.

For our function: \[ a_0 = \frac{1}{2} \bigg( \int_{0}^{1} x \, dx \bigg) = \frac{1}{2} \frac{x^2}{2} \bigg|_{0}^{1} = \frac{1}{4} \]
Thus, the constant term $$a_0$$ is 1/4.

Integration of piecewise functions involves breaking the integral into parts where the function definition changes.

For our function, the interval [-1, 1] is split into:

• From -1 to 0, where $$f(x) = 0$$
• From 0 to 1, where $$f(x) = x$$

By integrating each piece separately, we handle the discontinuity in the function. A piecewise function simplifies the integration as we only consider the contributions from intervals where the function isn't zero. Here’s the integration for our function during the region from 0 to 1:

\[ \int_{0}^{1} x \, dx = \frac{x^2}{2} \bigg|_{0}^{1} = \frac{1}{2} \]
In situations where the function is defined by multiple intervals, breaking it down and integrating each interval separately simplifies the calculation.

A Fourier series is a way to represent a function as a sum of sine and cosine terms. For our function, it's a series formed as:
\br> \[ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \]
For odd functions, like ours, the cosine terms ($$a_n$$) often turn out to be zero, leaving only the sine terms ($$b_n$$).

The sine coefficients are calculated using the given formula:

• \[ b_n = \int_{-L}^{L} f(x)\sin(n\pi x) \, dx \]

Since our function is only non-zero from 0 to 1, the integral for the sine coefficients reduces to:

\[ b_n = 2 \int_{0}^{1} x\sin(n\pi x) \, dx \]

Integration by parts is a method used to integrate products of functions. It's based on the formula:
\br> \[ \int u \, dv = uv - \int v \, du \]
To integrate the sine coefficients, we choose:

• $$u = x$$ and $$dv = \sin(n\pi x) \, dx$$

Therefore:

• $$du = dx$$
• $$v = -\frac{1}{n\pi} \cos(n\pi x)$$

Applying integration by parts, we get:
\br> \[ \int_0^1 x \sin(n\pi x) \, dx = \left[ -\frac{x \cos(n\pi x)}{n\pi} \right]_0^1 + \frac{1}{n\pi}\int_0^1 \cos(n\pi x) \, dx \]
Evaluating this and simplifying, we find the sine coefficient for each term of our series.

Fourier series can be used to solve differential equations, particularly those with periodic inputs. By representing the non-homogeneous term as a Fourier series, we can match coefficients to find the solution.

In our example, we start with:

• \[ y'' + 9y = f(x) \]

where we use the Fourier series representation of $$f(x)$$:

• \[ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left( (-1)^{n+1} \frac{2}{n\pi} \right) \sin(n\pi x) \]

We assume a solution form for $$y(x)$$:

• \[ y(x) = \sum_{n=1}^{\infty} C_n \sin(n\pi x) \]

Substituting $$y(x)$$ into the differential equation and equating coefficients, we solve for $$C_n$$.
This process links each sine term on the right-hand side to the respective term in the general solution, allowing us to find the particular solution.