Die folgenden Anfangswertprobleme beschreiben mechanische Schwingungen im aperiodischen Grenz fall. Wie lauten die L?sungen? a) \(2 \vec{x}+10 \dot{x}+12,5 x=0, \quad x(0)=5, \quad \dot{x}(0)=1\) b) \(\quad \ddot{x}+\dot{x}+0,25 x=0, \quad x(0)=1, \quad \dot{x}(0)=-1\)

Second-order differential equations are equations that involve the second derivative of a function. They are common in various fields, especially in physics where they often describe systems with acceleration, such as mechanical oscillations. A general form of a second-order differential equation with constant coefficients can be written as:
\[ a \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + c x = 0 \] Here, the equation is homogeneous because the right side is zero. This means there are no external forces acting on the system. The coefficients \(a\), \(b\), and \(c\) are constants, making it easier to solve systematically.

• \( \frac{d^2 x}{dt^2} \) represents the acceleration or second derivative of \(x\)
• \( \frac{dx}{dt} \) represents the velocity or first derivative of \(x\)
• \( x \) is the position or the function itself

In mechanical oscillations, these terms collectively describe the motion of the system over time.

The characteristic equation is a crucial step in solving second-order differential equations. It transforms the differential equation into an algebraic equation. For an equation of the form:
\[ a \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + c x = 0 \] We find the characteristic equation by assuming a solution of the form \(x = e^{rt}\), where \(r\) is a constant. Substituting \(x = e^{rt}\) and its derivatives into the differential equation, we get:
\[ a(r^2e^{rt}) + b(re^{rt}) + c(e^{rt}) = 0 \] Since \(e^{rt}\) is never zero, it simplifies to:
\[ ar^2 + br + c = 0 \] This is a quadratic equation in \(r\). The solutions to this quadratic equation are the roots \(r_1\) and \(r_2\), which determine the form of the general solution to the differential equation.

Initial conditions are values given for the function and its derivatives at a specific point, usually \( t = 0 \). These conditions allow us to find the specific solution or particular solution to a differential equation. Initial conditions are typically given in the form:
\[ x(0) = x_0 \quad \text{and} \quad \dot{x}(0) = v_0 \] Where \( x_0 \) is the initial position and \( v_0 \) is the initial velocity. Applying these conditions helps us determine the constants in our general solution. For example, if the general solution to a differential equation is:
\[ x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} \] We substitute \( t = 0 \) and the initial conditions to form two equations that can be solved for \( c_1 \) and \( c_2 \). This helps specify the unique solution that fits both the differential equation and the initial conditions.

The general solution to a second-order differential equation encompasses all possible solutions and typically involves two arbitrary constants. For a characteristic equation with roots \( r_1 \) and \( r_2 \):
\[ x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} \] If the roots are real and distinct, this is the general solution. If the roots are repeated, say \( r_1 = r_2 = r \), the general solution changes to:
\[ x(t) = (c_1 + c_2 t) e^{r t} \] The particular solution is found by applying the initial conditions to the general solution, giving us specific values for \( c_1 \) and \( c_2 \). For instance, if \( x(0) = x_0 \) and \( \dot{x}(0) = v_0 \):

• Substitute \( t = 0 \) into the general solution to get one equation.
• Differentiate the general solution and substitute \( t = 0 \) to get another equation.
• Solve these equations to find the constants \( c_1 \) and \( c_2 \).

Example:

Given the initial conditions are \( x(0) = 1 \) and \( \dot{x}(0) = -1 \), and using the general solution form:
\[ x(t) = (c_1 + c_2 t)e^{-\frac{1}{2}t} \] We substitute the initial conditions to find \(c_1\) and \(c_2\), giving us the particular solution that fits the initial setup.