Lösen Sie die folgenden Anfangswertprobleme durch Trennung der Variablen: a) \(y^{\prime}+(\cos x) \cdot y=0, \quad y\left(\frac{\pi}{2}\right)=2 \pi\) b) \(x(x+1) y^{\prime}=y, \quad y(1)=\frac{1}{2}\) c) \(y^{2} y^{\prime}+x^{2}=1, \quad y(2)=1\)

Anfangswertprobleme, also known as initial value problems, are a type of problem in differential equations where you need to determine a solution that satisfies a given initial condition. This involves finding a particular solution to a differential equation that passes through a specific point.

For example, in the exercise given:

• a) \(y^{\backslash prime}+(\backslash cos x) \backslash cdot y=0, \quad y\backslash left(\backslash frac{\backslash pi}{2}\backslash right)=2 \backslash pi\)
• b) \(x(x+1) y^{\backslash prime}=y, \quad y(1)=\backslash frac{1}{2}\)
• c) \(y^{2} y^{\backslash prime}+x^{2}=1, \quad y(2)=1\)

The goal here is to first solve the differential equation and then use the initial condition to find a specific solution. This extra step ensures your solution fits the particular initial value.

Differentialgleichungen, or differential equations, are equations that involve functions and their derivatives. These equations describe how one quantity changes with respect to another.

In the context of the problem, differential equations are used to model dynamically changing systems. For example:

• In equation (a), \(y^{\backslash prime}+(\backslash cos x) \backslash cdot y=0\) models a system where the rate of change of \(y\) depends on both \(y\) and \(x\).
• In equation (b), \(x(x+1) y^{\backslash prime}=y\), the rate of change of \(y\) is influenced by a combination of \(x\) and \(x+1\).
• In equation (c), \(y^{2} y^{\backslash prime}+x^{2}=1\), the rate of change of \(y\) is influenced by \(y^2\) and \(x^2\).

Solving these equations involves finding a function, or set of functions, that satisfy the relation stipulated by the differential equation.

Flächenintegration, or area integration, refers to the method of integrating a function over a region or area. In solving differential equations by separation of variables, we often use area integration to evaluate the integral on either side of the equation.

For instance, in the provided problems, area integration helps in breaking down the differential equations. When we separate the variables, we integrate each side with respect to its variable:

• For (a), \(\backslash int \backslash frac{1}{y} \, dy = -\backslash int \backslash cos x \, dx\)
• For (b), \(\backslash int \backslash frac{1}{y} \, dy = \int \backslash frac{1}{x(x+1)} \, dx\)
• For (c), \(\backslash int \backslash frac{1}{y^2} \, dy = \int \backslash frac{1-x^2}{1} \, dx\).

This approach makes it manageable to evaluate and solve for the integration constants.

Integration is a fundamental concept in calculus that involves finding the integral of a function. It represents the accumulation of quantities and is essentially the reverse process of differentiation.

Integrating after separating variables allows us to solve differential equations. For instance, after separating variables in problem (a) \( \backslash int \backslash frac{1}{y} \, dy = -\backslash int \backslash cos x \, dx\), we integrate each side:

• \( \backslash int \backslash frac{1}{y} \, dy = \backslash ln|y| + C_1 \)
• \(-\backslash int \backslash cos x \, dx = -\backslash sin x + C_2 \)

Finally, combining integrals and using initial conditions to solve the constants helps in finding the explicit solutions. These steps ensure a thorough understanding of the integration process involved in differential equations.