Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

Gegeben ist für \(x>0\) die Differentialgleichung $$ y^{\prime \prime}-\frac{2}{x} \cdot y^{\prime}+\frac{2}{x^{2}} \cdot y=0 $$ mit den linear unabhängigen Lösungen $$ y_{1}(x)=x, \quad y_{2}(x)=x^{2} $$ Untersuchen Sie die Randwertprobleme, die sich bei folgenden Randbedingungen ergeben, auf Lösbarkeit und bestimmen Sie gegebenenfalls die Lösungen: a) \(y(1)=5, \quad y(2)=16\), b) \(\quad y^{\prime}(1)=2, \quad y(2)=2\), c) \(\quad 2 \cdot y\left(\frac{1}{2}\right)-y^{\prime}\left(\frac{1}{2}\right)=2, \quad y(2)-2 \cdot y^{\prime}(2)=16\).

Step-by-Step Solution

Answer

For (a): The solution is \( y(x) = 2x + 3x^2 \). For (b) and (c): No solutions exist.

Step 1 - General Solution of the Differential Equation

The given differential equation is: \[ y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0 \] The general solution to such an equation is a linear combination of its linearly independent solutions. Since the solutions provided are \( y_1(x) = x \) and \( y_2(x) = x^2 \), the general solution is: \[ y(x) = C_1 x + C_2 x^2 \]

Step 2 - Boundary Conditions for (a)

Analyze the boundary conditions: \[ y(1) = 5, \quad y(2) = 16 \] Using \( y(x) = C_1 x + C_2 x^2 \), substitute the conditions: When \( x = 1 \), \( y(1) = C_1 + C_2 = 5 \). When \( x = 2 \), \( y(2) = 2C_1 + 4C_2 = 16 \).

Step 3 - Solve System of Equations for (a)

Solve the system: \[ C_1 + C_2 = 5 \] \[ 2C_1 + 4C_2 = 16 \] First equation: \( C_1 = 5 - C_2 \) Substitute in the second equation: \[ 2(5 - C_2) + 4C_2 = 16 \] Simplify: \[ 10 - 2C_2 + 4C_2 = 16 \] \[ 10 + 2C_2 = 16 \] \[ 2C_2 = 6 \] \[ C_2 = 3 \] Substitute \( C_2 \): \[ C_1 + 3 = 5 \] \[ C_1 = 2 \] Therefore, the particular solution is: \[ y(x) = 2x + 3x^2 \]

Step 4 - Boundary Conditions for (b)

Analyze the boundary conditions: \[ y'(1) = 2, \quad y(2) = 2 \] The derivative of the general solution is: \( y'(x) = C_1 + 2C_2 x \). Substitute the conditions: When \( x = 1 \), \( y'(1) = C_1 + 2C_2 = 2 \). When \( x = 2 \), \( y(2) = 2C_1 + 4C_2 = 2 \).

Step 5 - Solve System of Equations for (b)

Solve the system: \[ C_1 + 2C_2 = 2 \] \[ 2C_1 + 4C_2 = 2 \] The second equation is a multiple of the first, indicating a contradiction: \[ ((2C_1 + 4C_2)/2C_1) = 2 if 4C_2/2C_1=2 \] which is a contradiction yielding only zeros. Thus, there is no solution for these boundary conditions.

Step 6 - Boundary Conditions for (c)

Analyze the boundary conditions: \[ 2y(\frac{1}{2}) - y'(\frac{1}{2}) = 2, \quad y(2) - 2y'(2) = 16 \] Substitute the values: Using \( y(x) = C_1 x + C_2 x^2 \) and \( y'(x) = C_1 + 2C_2 x \): When \( x = \frac{1}{2} \), \( y(\frac{1}{2}) = \frac{1}{2}C_1 + \frac{1}{4}C_2 \) and \( y'(\frac{1}{2}) = C_1 + C_2 \). When \( x = 2 \), \( y(2) = 2C_1 + 4C_2 \) and \( y'(2) = C_1 + 4C_2 \).

Step 7 - Solve System of Equations for (c)

Substitute into the equations: \[ 2(\frac{1}{2}C_1 + \frac{1}{4}C_2) - (C_1 + C_2) = 2 \] Simplifies to \[ C_1 + \frac{1}{2} C_2 - C_1 + C_1 - \frac{1}{4} C_2 = 2 \] When \( C_1 = \frac{2 \times 4 - C_1}{C_2} \) Simplifies to: \[ y'(\frac{q}{1}) = C_2 - 2 \] Therefore, subtract: \[ \frac{2 - C_1' } = 1 \] The boundary conditions contradict. The systems have no solutions.

Key Concepts

differential equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. In simple terms, they show how a quantity changes over time or space. For example, the given differential equation is:
\[ y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0 \]
This equation involves the second derivative \(y''\) and first derivative \(y'\) of the function \(y\) with respect to the variable \(x\). Differential equations can be of various types, such as ordinary (involving a single variable) or partial (involving multiple variables). The primary goal is to find the function \(y\) that satisfies these relationships.

linearly independent solutions

Linearly independent solutions are essential in solving differential equations. They ensure that the combination of solutions provides a complete set of possible solutions. For the given differential equation, the provided linearly independent solutions are:
\( y_1(x) = x \) and \( y_2(x) = x^2 \)
These solutions are linearly independent if no constant multiple of one solution equals the other, i.e., \( y_1 eq k y_2 \) for some constant \(k\). They form the basis for the general solution of the differential equation. This concept helps us explore all possible solutions by combining these independent functions.

general solution

The general solution of a differential equation is a combination of its linearly independent solutions. For our equation, the general solution is:
\[ y(x) = C_1 x + C_2 x^2 \]
Here, \(C_1\) and \(C_2\) are arbitrary constants that we determine using given boundary conditions. This form represents all potential solutions to the differential equation. By adjusting \(C_1\) and \(C_2\), we can meet specific boundary conditions to find particular solutions tailored to the problem at hand.

boundary conditions

Boundary conditions specify the values of the function or its derivatives at specific points. These conditions help narrow down the general solution to a particular solution. Let's see three scenarios from our problem:

• For \( y(1) = 5 \) and \( y(2) = 16 \), we substituted these into the general solution to find specific values of \( C_1 \) and \( C_2 \).
• For \( y'(1) = 2 \) and \( y(2) = 2 \), the system of equations presented a contradiction, indicating no solution.
• For complicated conditions like \( 2y(\frac{1}{2}) - y'(\frac{1}{2}) = 2 \) and \( y(2) - 2y'(2) = 16 \), we also found contradictions.

Boundary conditions are crucial in finding specific solutions that fit the context of a problem.