Let's break down and solve the mathematical problems one by one from the image you uploaded.

1. Bestimmen Sie die Lösung des AWP:

$(2x+1)y'' - 3y = 0 \quad \text{mit} \quad y(0) = 24, \, y'(0) = 0$

This is a second-order linear differential equation. We'll use the method of a power series to find the solution.

Step 1: Power Series Ansatz

The problem suggests using a power series ansatz:

$g(x) = \sum_{n=0}^{\infty} a_n x^n$

Taking the second derivative:

$g'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad g''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

Substitute these into the differential equation:

$(2x+1)g''(x) - 3g(x) = 0$

Expanding the terms:

$(2x+1)\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 3\sum_{n=0}^{\infty} a_n x^n = 0$

Multiplying out $(2x+1)g''(x)$:

$2x\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

Combine terms by shifting indices where necessary and solve for the recurrence relation of the coefficients $a_n$.

Step 2: Recurrence Relation

The resulting recurrence relation is:

$a_{n+2} = \frac{3a_n - 2n a_n}{(n+2)(n+1)}$

This gives a formula to compute all coefficients based on $a_0$ and $a_1$.

Step 3: Initial Conditions

Using the initial conditions $y(0) = 24$ and $y'(0) = 0$, we set:

$a_0 = 24, \quad a_1 = 0$

Step 4: Explicit Calculation of Coefficients

We can now calculate the first few coefficients:

$a_2 = \frac{3a_0 - 2 \times 0 \times a_0}{2 \times 1} = \frac{3 \times 24}{2} = 36$

$a_3 = 0 \quad (\text{since} \, a_1 = 0)$

$a_4 = \frac{3a_2 - 2 \times 2 \times a_2}{4 \times 3} = \frac{3 \times 36 - 4 \times 36}{12} = -3$

Continue this process to calculate $a_5$ and beyond.

2. Fourier Series:

The function $g(t) = t^2 + |t|$ is given, and we are asked to compute the Fourier coefficients.

The function is $2\pi$-periodic, and we have:

$b_k = 0, \quad \text{for all} \, k$

Step 1: Fourier Coefficient $a_0$

The constant term $a_0$ is given by:

$a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \, dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} (t^2 + |t|) \, dt$

This integral can be split at $t = 0$, and the values can be computed accordingly.

Step 2: Fourier Coefficients $a_k$

The Fourier coefficients $a_k$ are given by:

$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} g(t) \cos(kt) \, dt$

For this particular function, you can compute the integral of $t^2 \cos(kt)$ and $|t| \cos(kt)$ separately.

Would you like more details on each step or have any specific questions about the method? Here are 5 related questions:

1. How do we handle non-homogeneous terms in differential equations using a power series?
2. Why do Fourier coefficients $b_k$ vanish for even functions?
3. How does the choice of the initial conditions $y(0)$ and $y'(0)$ affect the power series solution?
4. Can we approximate the solution of a second-order ODE numerically using a power series?
5. How can symmetry properties of $g(t)$ simplify the calculation of Fourier coefficients?

Tip: In many cases, calculating the first few coefficients in a power series gives a good approximation of the solution for small $x$.