Problem 5 - Arbeitsbuch Mathematik für Ingenieure

Question

Überprüfen Sie, ob die angegebenen Funktionen auf dem Intervall \((0, \infty)\) ein Fundamentalsystem der Differentialgleichung $$ y^{\prime \prime}-\frac{x+2}{x} \cdot y^{\prime}+\frac{1}{x} \cdot y=0 $$ bilden: a) \(y_{1}(x)=e^{x} \cdot(x-2), \quad y_{2}(x)=3 x+6\) b) \(y_{1}(x)=e^{x} \cdot(x-2), \quad y_{3}(x)=x-3\), c) \(y_{1}(x)=e^{x} \cdot(x-2), \quad y_{4}(x)=4+2 x-2 e^{x}+x e^{x}\) Bestimmen Sie nun diejenige Lösung der obigen Differentialgleichung, die die Anfangsbedingungen $$ y(2)=8, \quad y^{\prime}(2)=2+e^{2} $$ erfüllt.

Step-by-Step Solution

Answer

The functions in part (a) and (c) form a fundamental system. The particular solution that meets the initial conditions is \(y(x) = e^x (x-2) + 2x + 4\).

Step 1: Verify Linear Independence

To determine if the given functions form a fundamental system of the differential equation, check if they are linearly independent. Use the Wronskian determinant for the pairs of functions in parts (a), (b), and (c). The Wronskian of functions \(y_1(x)\) and \(y_2(x)\) is given by:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix}\]

Step 2: Calculate Wronskian for Part (a)

For part (a), the functions are \(y_1(x) = e^{x} (x-2)\) and \(y_2(x) = 3x + 6\). Calculate their derivatives:\(y_1'(x) = (x-1)e^x\) and \(y_2'(x) = 3\).Then compute the Wronskian:\[W(y_1, y_2) = \begin{vmatrix} e^{x} (x-2) & 3x+6 \ (x-1)e^x & 3 \end{vmatrix}= e^{x} (x-2) \times 3 - (3x+6)(x-1)e^x = -3e^x eq 0\]Since the Wronskian is non-zero, \(y_1(x)\) and \(y_2(x)\) are linearly independent on the interval \((0, \, \infty)\).

Step 3: Calculate Wronskian for Part (b)

For part (b), the functions are \(y_1(x) = e^{x} (x-2)\) and \(y_3(x) = x-3\). Calculate their derivatives:\(y_1'(x) = (x-1)e^x\) and \(y_3'(x) = 1\).Then compute the Wronskian:\[W(y_1, y_3) = \begin{vmatrix} e^{x} (x-2) & x-3 \ (x-1)e^x & 1 \end{vmatrix}= e^{x} (x-2) \times 1 - (x-3)(x-1)e^x = 0\]Since the Wronskian is zero, \(y_1(x)\) and \(y_3(x)\) are linearly dependent on the interval \((0, \, \infty)\).

Step 4: Calculate Wronskian for Part (c)

For part (c), the functions are \(y_1(x) = e^{x} (x-2)\) and \(y_4(x) = 4+2x-2e^x+xe^x\). Calculate their derivatives:\(y_1'(x) = (x-1)e^x\) and \(y_4'(x) = 2 + e^x + xe^x\).Then compute the Wronskian:\[W(y_1, y_4) = \begin{vmatrix} e^{x} (x-2) & 4+2x-2e^x+xe^x \ (x-1)e^x & 2 + e^x + xe^x \end{vmatrix}= -2e^{2x} eq 0\]Since the Wronskian is non-zero, \(y_1(x)\) and \(y_4(x)\) are linearly independent on the interval \((0, \, \infty)\).

Step 5: Find General Solution of Differential Equation

From part (a), the general solution to the differential equation on the interval \((0, \, \infty)\) using \(y_1(x)\) and \(y_2(x)\) is:\[ y(x) = C_1 e^{x} (x-2) + C_2 (3x + 6) \]

Step 6: Apply Initial Conditions

Use the initial conditions \(y(2) = 8\) and \(y'(2) = 2 + e^2\). First, calculate\[ C_1 e^{2}(2-2) + C_2(3 \cdot 2 + 6) = 8 \quad \Rightarrow \quad 12C_2 = 8 \quad \Rightarrow \quad C_2 = \frac{2}{3} \]Next, find the derivative:\[ y'(x) = C_1 ((x-1)e^x) + C_2 \cdot 3 \quad \Rightarrow \quad y'(2) = C_1 \cdot e^2 + \frac{2}{3} \cdot 3 = 2 + e^2 \quad \Rightarrow \quad C_1 e^2 + 2 = 2 + e^2 \quad \Rightarrow \quad C_1 = 1 \]

Step 7: Write Particular Solution

Using the values of \(C_1\) and \(C_2\) found, the particular solution is:\[ y(x) = 1 \cdot e^{x} (x-2) + \frac{2}{3} (3x + 6) = e^x (x-2) + 2x + 4\]

Key Concepts

Wronskian

The Wronskian is a determinant used to test whether a set of functions are linearly independent. To calculate it for functions \( y_1(x) \) and \( y_2(x) \), use the formula:

\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]
It involves first finding the derivatives of the functions. For example, for functions \( y_1(x) = e^x (x-2) \) and \( y_2(x) = 3x + 6 \), their derivatives are \( y_1' = (x-1)e^x \) and \( y_2' = 3 \). Plug these into the matrix and compute the determinant:
\[ W(y_1, y_2) = e^{x} (x-2) \cdot 3 - (3x+6)(x-1)e^x = -3e^x \]
If the Wronskian is non-zero for some interval, the functions are linearly independent over that interval.

Linear Independence

Linear independence implies that no function in a set can be written as a linear combination of others in the set. To check if functions \( y_1(x) \) and \( y_2(x) \) are linearly independent, compute their Wronskian. If it is non-zero, they are linearly independent.
For instance, with \( y_1(x) = e^x (x-2) \) and \( y_2(x) = 3x + 6 \), their Wronskian is \( -3e^x \). Since this is not zero, they are linearly independent on the interval \( (0, \, \infty) \).
If the Wronskian equals zero, such as for \( y_1(x) = e^x (x-2) \) and \( y_3(x) = x-3 \) where:
\[ W(y_1, y_3) = 0 \]
they are linearly dependent.

Particular Solution

A particular solution of a differential equation satisfies both the differential equation itself and the given initial conditions. After finding the general solution, plug in the initial conditions to solve for the constants.
The general solution of the equation \( y'' - \frac{x+2}{x} y' + \frac{1}{x} y = 0 \) using \( y_1(x) = e^x (x-2) \) and \( y_2(x) = 3x + 6 \) is:
\[ y(x) = C_1 e^x (x-2) + C_2 (3x + 6) \]
Given the initial conditions \( y(2) = 8 \) and \( y'(2) = 2 + e^2 \), find the constants \( C_1 \) and \( C_2 \). First, use \( y(2) = 8 \) to find \( C_2 \):
\[ 12C_2 = 8 \quad \Rightarrow \quad C_2 = \frac{2}{3} \]
Then, use \( y'(2) = 2 + e^2 \) to find \( C_1 \):
\[ C_1 e^2 + 2 = 2 + e^2 \quad \Rightarrow \quad C_1 = 1 \]
The particular solution is thus:
\[ y(x) = e^x (x-2) + 2x + 4 \]

Initial Conditions

Initial conditions provide specific values for the solution of a differential equation at a particular point. They are essential for finding a particular solution from a general solution. For the differential equation:
\[ y'' - \frac{x+2}{x} y' + \frac{1}{x} y = 0 \]
find the general solution first, which in our case is:
\[ y(x) = C_1 e^x (x-2) + C_2 (3x + 6) \]
Given the initial conditions \( y(2) = 8 \) and \( y'(2) = 2 + e^2 \), use these values to find \( C_1 \) and \( C_2 \). Start by substituting \( x = 2 \) in the general solution to get:
\[ C_1 e^2 (0) + C_2 (12) = 8 \quad \Rightarrow \quad C_2 = \frac{2}{3} \]
Next, differentiate the general solution and use \( y'(2) = 2 + e^2 \) to find \( C_1 \):
\[ C_1 e^2 + 2 = 2 + e^2 \quad \Rightarrow \quad C_1 = 1 \]
This gives us the particular solution satisfying initial conditions:
\[ y(x) = e^x (x-2) + 2x + 4 \]