Let's break down the problem based on the image you provided:

Task 1:

We are given the second-order differential equation: $(2x + 1) y'' - 3y = 0$ with initial conditions $y(0) = 24$ and $y'(0) = 0$. We are to use a power series method to solve this initial value problem (AWP in German stands for "Anfangswertproblem").

We assume a power series solution of the form: $y(x) = \sum_{n=0}^{\infty} a_n x^n$ which implies that its derivatives are: $y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}$ $y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

a) Find the recurrence relation:

Substituting the power series for $y(x)$ and its derivatives into the given differential equation, we aim to derive a recurrence relation for $a_n$. We will also calculate the first few coefficients.

b) Calculate the explicit values for $a_0, a_1, a_2, a_3, a_4, a_5$:

We are given $a_0 = 24$, $a_1 = 0$, and from the recurrence relation we calculate the next coefficients.

Task 2:

We are given the 2π-periodic extension of the function: $g(t) = t^2 + |t| \quad \text{for} \quad -\pi < t \leq \pi$ and need to calculate the real Fourier coefficients $a_0$, $a_k$, and $b_k$.

The function is even, so all $b_k = 0$, and we only need to compute the Fourier coefficients $a_0$ and $a_k$.

Let me proceed by calculating the recurrence relation for Task 1(a) and the Fourier coefficients for Task 2. I'll perform these calculations step by step.We have expanded the equation $(2x + 1)y'' - 3y = 0$, and the result involves sums of powers of $x$. The next step is to collect terms in powers of $x$ and derive the recurrence relation for the coefficients $a_n$. This will give us the necessary recurrence formula.

Let me proceed with that simplification to extract the recurrence relation explicitly.The simplified form still includes terms of sums with powers of $x$. We will now equate the coefficients of like powers of $x$ to derive the recurrence relation for $a_n$. This will give us a formula for $a_{n+2}$ in terms of previous coefficients. Let's proceed with that step.It seems that the recurrence relation simplifies to $a_n = 0$, which suggests that higher-order terms beyond a certain degree may vanish in this specific case. However, the non-trivial values for $a_0$ and $a_1$ provide the initial conditions for the power series expansion.

Let's now calculate the first few coefficients explicitly for $a_0, a_1, a_2, a_3, a_4,$ and $a_5$, based on the recurrence relation and the initial conditions provided in the problem.It seems like I can't do more advanced data analysis right now. Please try again later.

However, based on the image, the values for the first few coefficients are:

• $a_0 = 24$
• $a_1 = 0$
• $a_2 = 36$
• $a_3 = -24$
• $a_4 = 33$
• $a_5$ remains to be calculated or seems to be missing from the provided hint.

Let me know if you'd like further clarification or help with other parts of the problem.

Here are 5 related questions to deepen understanding:

1. What steps are involved in solving differential equations using power series methods?
2. How do initial conditions affect the solution of a differential equation?
3. What is the importance of recurrence relations in power series solutions?
4. How does Fourier series help represent periodic functions like $g(t) = t^2 + |t|$?
5. How would you calculate the remaining Fourier coefficients for the function in Task 2?

Tip: When solving differential equations with power series, always verify the initial conditions by plugging them back into the original equation.