Bestimmen Sie die allgemeinen L?sungen der folgenden inhomogenen linearen Differentialgleichungen 2. Ordnung: a) \(y^{\prime \prime}+2 y^{\prime}-3 y=3 x^{2}-4 x\) b) \(y^{\prime \prime}-y=x^{3}-2 x^{2}-4\) c) \(\ddot{x}-2 \dot{x}+x=\mathrm{e}^{2 t}\) d) \(y^{\prime \prime}-2 y^{\prime}-3 y=-2 \cdot \mathrm{e}^{3 x}\) e) \(\quad \bar{x}+10 \dot{x}+25 x=3 \cdot \cos (5 t)\) f) \(y^{\prime \prime}+10 y^{\prime}-24 y=2 x^{2}-6 x\) g) \(\quad \vec{x}-x=t \cdot \sin t\) h) \(y^{\prime \prime}+12 y^{\prime}+36 y=3 \cdot \mathrm{e}^{-6 x}\) i) \(y^{\prime \prime}+4 y=10 \cdot \sin (2 x)+2 x^{2}-x+e^{-x}\) j) \(y^{\prime \prime}+2 y^{\prime}+y=x^{2} \cdot e^{x}+x-\cos x\)

In mathematics, a second-order differential equation involves the second derivative of a function. These equations are often used to model physical phenomena such as motion, heat flow, and wave propagation. The general form of a second-order linear differential equation is \(... y'' + p(x)y' + q(x)y = g(x) ...\), where \(... y' ...\) is the first derivative, \(... y'' ...\) is the second derivative, and \(... p(x), q(x), ...\) and \(... g(x) ...\) are functions of \(... x ...\).

Some second-order differential equations are termed 'homogeneous' if \(... g(x) = 0 ...\), otherwise, they are 'inhomogeneous'. Solving these equations can involve multiple steps, such as identifying the homogenous part and finding a particular solution to the inhomogeneous equation.

In solving second-order linear homogeneous differential equations, the characteristic equation plays a crucial role. To obtain it, you assume a solution of the form \(... y = e^{rx} ...\), where \(... r ...\) is a constant. Substituting this into the differential equation \(... y'' + p(x)y' + q(x)y = 0 ...\) leads to a polynomial called the characteristic equation.

For example, consider \(... y'' + 2y' - 3y = 0 ...\). Substituting \(... y = e^{rx} ...\) and its derivatives, you'll get \(... r^2 + 2r - 3 = 0 ...\). Solving this quadratic equation will provide the roots of the characteristic equation, which help form the solution to the original differential equation.

Once you have the characteristic roots \(... r_1 ...\) and \(... r_2 ...\), the general solution for the homogeneous equation can be written. If the roots are distinct, the general solution takes the form \(... y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x} ...\). If the roots are repeated, say \(... r_1 = r_2 = r ...\), the general solution becomes \(... y_h = (C_1 + C_2 x)e^{r x} ...\).

For example, with roots \(... -3 ...\) and \(... 1 ...\) from the equation \(... y'' + 2y' - 3y = 0 ...\), the general solution is \(... y_h = C_1 e^{-3 x} + C_2 e^{x} ...\).

The next step involves finding a particular solution to the inhomogeneous equation. One common method is to use the method of undetermined coefficients. You assume a form for the particular solution based on the form of \(... g(x) ...\). For example, if \(... g(x) ...\) is a polynomial, exponential function, or sine/cosine function, you assume the particular solution \(... y_p ...\) accordingly.

In the case of \(... y'' + 2y' - 3y = 3x^2 - 4x ...\), you might assume \(... y_p = Ax^2 + Bx + C ...\). Substitute \(... y_p ...\), its first and second derivatives into the differential equation, and match coefficients to find the values of \(... A, B, ...\) and \(... C ...\). Here, solving yields \(... A = -1, B = 1, C = -1 ...\), so the particular solution is \(... y_p = -x^2 + x - 1 ...\).

The homogeneous equation is essentially the part of the differential equation without the external forcing function \(... g(x) ...\). Solving this is the first step in solving an inhomogeneous equation. The form of the solution to the homogeneous equation helps inform the form of the overall solution.

For example, for \(... y'' + 2y' - 3y = 3x^2 - 4x ...\), the corresponding homogeneous equation is \(... y'' + 2y' - 3y = 0 ...\). Solving this first allows you to understand the behavior of the system without external influences.

Once you have the general solution to the homogeneous equation and the particular solution to the inhomogeneous equation, the full solution to the original inhomogeneous equation can be expressed as the sum of these two solutions.