L?sen Sie die folgenden Anfangswertprobleme: a) \(y^{\prime \prime \prime}+9 y^{\prime}=18 x, \quad y(\pi)=\pi^{2}, \quad y^{\prime}(\pi)=2 \pi, \quad y^{\prime \prime}(\pi)=20\) b) \(y^{\prime \prime \prime}+8 y^{\prime \prime}+17 y^{\prime}+10 y=34 \cdot \sin x+12 \cdot \cos x\) \(y(0)=1, \quad y^{\prime}(0)=-3, \quad y^{\prime \prime}(0)=8\) c) \(x^{(4)}-x=45 \cdot \mathrm{e}^{2 t}, \quad x(0)=6, \quad \dot{x}(0)=0, \quad \ddot{x}(0)=15, \quad \dddot x(0)=24\) d) \(v^{(5)}-\dot{v}=2 t+2\) $$ v(0)=1, \quad \dot{v}(0)=-2, \quad \ddot{v}(0)=2, \quad \dddot v(0)=0, \quad v^{(4)}(0)=-4 $$

A higher-order differential equation is one that involves derivatives of an unknown function of orders two or higher. Typically, these equations are written in the form: d^n y/dx^n + a_{n-1} d^(n-1) y/dx^(n-1) + ... + a_1 dy/dx + a_0 y = g(x), where n is the order of the differential equation, and a_0, a_1, ..., a_(n-1) are constants or functions of x.
Unlike first-order differential equations, solving higher-order equations typically involves finding both the homogeneous solution (solution to the associated homogeneous equation) and a particular solution to the non-homogeneous equation.
Understanding higher-order differential equations is essential to many fields like physics, engineering, and economics.

The homogeneous solution of a differential equation is found by setting the right-hand side of the equation to zero (g(x) = 0) and solving: d^n y/dx^n + a_{n-1} d^(n-1) y/dx^(n-1) + ... + a_1 dy/dx + a_0 y = 0.
In such cases, we find the characteristic equation which is typically: r^n + a_{n-1} r^(n-1) + ... + a_1 r + a_0 = 0.
The roots (r) of the characteristic equation determine the form of the homogeneous solution. If the roots are real and distinct, the solution will be a combination of exponential terms. If they are complex, the solution will involve sines and cosines. For instance, in exercise part (a), the characteristic equation is: r^3 + 9r = 0, which factors to r(r^2 + 9) = 0, giving roots r = 0, ±3i. Thus, the homogeneous solution is: y_h = C_1 + C_2 cos(3x) + C_3 sin(3x).

A particular solution to a non-homogeneous differential equation is a specific solution that fits the entire equation, including the non-homogeneous part (g(x)). To find it, we assume a form for y_p that depends on the nature of g(x).
Common forms include a polynomial, exponential, or trigonometric functions. For example, in exercise part (a), we assumed a polynomial form: y_p = Ax + B, because the right-hand side is 18x. After substituting this form into the differential equation and solving for constants, we found that A = 2.
So, y_p = 2x + B. Summing the particular and homogeneous solutions gives the general solution.

The characteristic equation is derived from the homogeneous version of the differential equation where the solution involves exponential functions. One sets up the polynomial equation in terms of r, using coefficients from the original differential equation.
For instance, in part (a) of the exercise: y''' + 9y' = 0, we take the characteristic polynomial r^3 + 9r = 0. Solving this equation yields the roots r = 0, ±3i. These roots determine the form of the homogeneous solution. Real roots correspond to exponential solutions, while complex roots lead to sine and cosine terms. Roots with multiplicity require adding polynomial terms to the solution.

Initial conditions are values provided for the function and its derivatives at a specific point, typically to solve for the constants in the general solution.
In the exercise, part (a) presented the initial conditions: y(π) = π^2, y'(π) = 2π, and y''(π) = 20. These conditions are used to find the constants C_1, C_2, and C_3 in the general solution: y = C_1 + C_2 cos(3x) + C_3 sin(3x) + 2x + B. Substituting the initial conditions into the general solution and its derivatives allows us to solve a system of equations to find the constants, ensuring the solution fits the specific problem context.