Bestimmen Sie die allgemeinen L?sungen der folgenden homogenen linearen Differentialgleichungen 5 . Ordnung: a) \(x^{(5)}+2 \dddot x+\dot{x}=0\) b) \(y^{(5)}+2 y^{(4)}+y^{\prime \prime \prime}=0\) c) \(y^{(5)}+3 y^{(4)}+10 y^{\prime \prime \prime}+6 y^{\prime \prime}+5 y^{\prime}-25 y=0\) d) \(y^{(5)}+22 y^{\prime \prime \prime}+2 y^{\prime \prime}-75 y^{\prime}+50 y=0\)

In solving homogeneous linear differential equations, the first crucial step is forming the characteristic equation. This equation helps simplify the problem significantly. To form it, replace each derivative in the differential equation with corresponding powers of a variable, typically denoted by 'r'. For example, a term like \(x^{(5)}\) would be replaced with \(r^5\), \(\frac{d^3 x}{dt^3}\) with \(r^3\), and \(\dot{x}\) with \(r\). This transformation converts a differential equation into a polynomial equation. Let's see an example with the equation \(x^{(5)} + 2 \dddot x + \dot{x} = 0\). The characteristic equation would be \(r^5 + 2r^3 + r = 0\). Once in this form, you can solve it like any other polynomial equation to find 'r', or the roots.

After forming the characteristic equation, the next step is to find its roots. These roots can be real or complex numbers and are crucial for determining the general solution of the differential equation. To find the roots, you might need to factorize the polynomial. For instance, the characteristic equation \(r^5 + 2r^3 + r = 0\) can be factored to \(r(r^2 + 1)^2 = 0\), giving us \(r = 0\) (root with multiplicity 1) and \(r = ±i\) (each with multiplicity 2). Solving these correctly is key to constructing the final solution of the differential equation.

Once the roots of the characteristic equation are found, they can be used to form the general solution of the differential equation. The nature (real vs. complex) and multiplicity (how many times a root appears) of each root guides the structure of the general solution. For example, if a root \(r\) has multiplicity \(m\), it generates \(m\) linearly independent solutions of the form \(e^{rt}\), \(t e^{rt}\), \(t^2 e^{rt}\), ..., \(t^{m-1} e^{rt}\). If the roots are complex, you will need to use Euler’s formula to express the solutions in terms of sine and cosine functions. For the equation \( x^{(5)} + 2 \dddot x + \dot{x} = 0 \), with roots \( r = 0 \) (multiplicity 1) and \( r = ±i \) (each with multiplicity 2), the general solution is \( x(t) = C_1 + (C_2 + C_3 t) e^{it} + (C_4 + C_5 t) e^{-it} \).

Complex roots occur when the characteristic equation has non-real solutions. These roots come in conjugate pairs like \(r = a ± bi\). When dealing with complex roots, we use Euler’s formula, which states that \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). Thus, for complex roots \(a ± bi\), the general solution will involve terms with sine and cosine. For example, if we have roots \(±i\), they translate into the solutions \(e^{it}\) and \(e^{-it}\), which using Euler’s formula becomes \( \cos(t) + i\sin(t)\) and its conjugate. In the general solution, these will take the form \((C_2 + C_3 t)e^{it} + (C_4 + C_5 t)e^{-it}\).

The multiplicity of a root refers to how many times a particular root appears in the characteristic equation. This is important because it affects the structure of the general solution. If a root \(r\) appears with multiplicity \(m\), it results in \(m\) linearly independent solutions of the form \(e^{rt}\), \(t e^{rt}\), ..., \(t^{m-1} e^{rt}\). For example, suppose the root \(r=0\) has a multiplicity of 3. Then the linearly independent solutions would be \(1, t, t^2\), leading to a general solution \(y(t) = C_1 + C_2 t + C_3 t^2 + ...\). This method makes sure that we account for the contribution of each root appropriately to form a complete solution.