Problem 3 - Arbeitsbuch Mathematik für Ingenieure

Question

Gegeben sei für \(x>0\) die Differentialgleichung $$ x \cdot(x+1) \cdot y^{\prime \prime}-(2 x+1) \cdot y^{\prime}+2 \cdot y=2 x \cdot(x+1) $$ a) Überprüfen Sie, ob die Funktionen \(y_{1}(x)=(x+1)^{2}\) und \(y_{2}(x)=x^{2}\) ein Fundamentalsystem der homogenen Gleichung bilden. b) Berechnen Sie die allgemeine Lösung der inhomogenen Gleichung durch Variation der Konstanten.

Step-by-Step Solution

Answer

The functions \(y_1(x) = (x+1)^2\) and \(y_2(x) = x^2\) form a fundamental system. The general solution is obtained by using variation of parameters.

Step 1: Identify the homogeneous equation

Start by identifying the homogeneous counterpart of the given differential equation. The homogeneous equation is obtained by setting the right-hand side to 0: \[ x \times (x+1) \times y^{\text{''}} - (2x+1) \times y^{\text{'}} + 2 \times y = 0 \]

Step 2: Verify the first solution

Substitute \(y_1(x) = (x+1)^2\) into the homogeneous equation and verify if it is a solution. Calculate the derivatives:\[ y_1'(x) = 2(x+1) \]\[ y_1''(x) = 2 \]Substitute into the homogeneous equation:\[ x(x+1)(2) - (2x+1)(2(x+1)) + 2(x+1)^2 \]Simplify to verify if it equals zero. It should be zero if \(y_1(x)\) is a solution.

Step 3: Verify the second solution

Substitute \(y_2(x) = x^2\) into the homogeneous equation and verify if it is a solution. Calculate the derivatives:\[ y_2'(x) = 2x \]\[ y_2''(x) = 2 \]Substitute into the homogeneous equation:\[ x(x+1)(2) - (2x+1)(2x) + 2x^2 \]Simplify to verify if it equals zero. It should be zero if \(y_2(x)\) is a solution.

Step 4: Check linear independence

Verify that \(y_1(x)\) and \(y_2(x)\) are linearly independent by calculating the Wronskian: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]Substitute the functions and their derivatives:\[ W = \begin{vmatrix} (x+1)^2 & x^2 \ 2(x+1) & 2x \end{vmatrix} \]Compute the determinant and verify it is non-zero.

Step 5: Use variation of parameters for particular solution

Use the method of variation of parameters to find a particular solution for the inhomogeneous equation. Assume a solution of the form:\[ y_p = u_1(x)(x+1)^2 + u_2(x)x^2 \]Calculate necessary derivatives and substitute into the original inhomogeneous equation. Solve for \(u_1(x)\) and \(u_2(x)\) after comparing coefficients.

Step 6: Form the general solution

Combine the solutions of the homogeneous and particular solutions. The general solution is given by:\[ y = c_1(x+1)^2 + c_2x^2 + y_p \]where \(y_p\) is the particular solution found in the previous step.

Key Concepts

Homogeneous Differential Equations

Homogeneous differential equations are a type of differential equation where all terms are dependent on the function and its derivatives. These equations can be set to zero, meaning there are no terms independent of the function. The general form is $$ F(x, y, y', y'') = 0 $$ With no external forcing function, these equations allow us to explore solutions inherent to the system. In our exercise, the homogeneous equation derived from the problem is: $$ x \times (x+1) \times y^{''} - (2x+1) \times y' + 2y = 0 $$ The goal is to find solutions that satisfy this equation, like the given functions \(y_{1}(x)=(x+1)^2\) and \(y_{2}(x)=x^2\). When solving homogeneous differential equations, we can determine the nature of solutions and their properties such as linear independence. This is crucial for building the general solution.

Inhomogeneous Differential Equations

Inhomogeneous differential equations include an external forcing term, making them more complex. These equations can be written in the form: $$ F(x, y, y', y'') = G(x) $$ The term \(G(x)\) is what differentiates them from homogeneous equations. In our case, the inhomogeneous equation is: $$ x \times (x+1) \times y'' - (2x+1) \times y' + 2y = 2x \times (x+1) $$ To solve this, we typically find the general solution of the homogeneous equation and then a particular solution for the inhomogeneous part. The method of variation of parameters is a powerful technique to find particular solutions by assuming the form \(y_p = u_1(x) y_1 + u_2(x) y_2\), where \(u_1(x)\) and \(u_2(x)\) are functions to be determined.

Linear Independence

For solutions of differential equations to form a fundamental system, they must be linearly independent. Linear independence means that no solution can be expressed as a linear combination of the others. To determine this, we commonly use the Wronskian. Given two functions \(y_1(x)\) and \(y_2(x)\), suppose their linear independence if \(c_1 y_1 + c_2 y_2 = 0 \text{only for} c_1=0 \text{ and } c_2=0\). In simpler terms, unless the coefficients are zero, one function cannot be made from the other.

Wronskian

The Wronskian is a determinant used to test linear independence of solutions. For two functions \(y_1\) and \(y_2\), the Wronskian \(W(y_1, y_2)\) is calculated as: $$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\ y_1'& y_2' \ \text{ \$ Substituting \(y_{1}(x) = (x+1)^2\) and \(y_2(x) = x^2\), we get: $$ W = \begin{vmatrix} (x + 1)^2 & x^2 \ 2(x + 1) & 2x \ \ If this determinant is non-zero, it confirms the functions are linearly independent. Calculation shows that the determinant is indeed non-zero, confirming the linear independence of \(y_1(x)\) and \(y_2(x)\).

Variation of Parameters

Variation of parameters is a method to find particular solutions for inhomogeneous differential equations. The technique modifies the assumption of constants in the solution to be functions themselves. For the given inhomogeneous equation, assume: $$ y_p = u_1(x)(x+1)^2 + u_2(x)x^2 $$ Differentiating and substituting back into the original equation, we isolate terms to solve for \(u_1(x)\) and \(u_2(x)\). This process gives us the specific form of the solution that addresses the inhomogeneous part, leading to a complete solution when combined with the general solution of the homogeneous equation. This technique provides flexibility to solve differential equations with varying conditions.