Problem 4 - Arbeitsbuch Mathematik für Ingenieure

Question

Gegeben sei die Differentialgleichung $$ y^{(5)}+y^{(4)}+y^{(3)}+y^{(2)}=b(x) . $$ a) Bestimmen Sie die allgemeine reelle Lösung durch Berechnung eines Fundamentalsystems der homogenen Gleichung und einer partikulären inhomogenen Lösung \(y^{*}(x)\) durch Ansatz vom Typ der Störfunktion für \(b(x)=2+e^{2 x}\). b) Geben Sie den Ansatz für \(y^{*}(x)\) an für die Störfunktion \(b(x)=\cos x+x \cdot e^{-x}\) (die explizite Berechnung von \(y^{*}(x)\) ist nicht verlangt).

Step-by-Step Solution

Answer

Solve the homogeneous equation's characteristic polynomial, combine homogeneous and particular solutions. The schosen ansatz for b(x)=cos x + x * e^{-x)

Step 1: Homogeneous Equation

Solve the homogeneous differential equation y^{(5)} + y^{(4)} + y^{(3)} + y^{(2)} = 0by finding the characteristic equation. The characteristic equation is obtained by assuming the solutions are of the form y(x) = e^{rx}, leading to the polynomial r^5 + r^4 + r^3 + r^2 = 0.

Step 2: Solve Characteristic Equation

Factor the characteristic polynomial: r^2(r^3 + r^2 + r + 1) = 0. Solve for r. The solutions are r_1=0 (double root) and the roots of the cubic equation r^3 + r^2 + r + 1 = 0. Use the Rational Root Theorem or numerical methods to solve the cubic equation if necessary.

Step 3: Fundamental System of Homogeneous Equation

The roots determine the fundamental solutions of the homogeneous equation. For each root r, the fundamental solution is of the form e^{rx}. Therefore, the general solution of the homogeneous equation is y_h(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} + C_3 e^{r_3 x} + C_4 e^{r_4 x}.Given the roots, write the specific form of this solution.

Step 4: Particular Solution for b(x)=2+e^{2x}

To find a particular solution y^∗(x), use the method of undetermined coefficients. Try a particular solution of the form y^∗(x) = A + B e^{2x}. Differentiate y^∗(x) and substitute into the differential equation to solve for the constants A and B.

Step 5: Substitute and Solve for Constants

First, differentiate y^∗(x) to get y^∗'(x) = 2B e^{2x}. y^∗''(x) = 4B e^{2x}, y^∗'''(x) = 8B e^{2x}, y^∗^(4)(x) = 16B e^{2x}, y^∗^(5)(x) = 32B e^{2x}.Substitute into the differential equation: 32B e^{2x} + 16B e^{2x} + 8B e^{2x} + 4B e^{2x} = 2 + e^{2x}. Separate e^{2x} terms and constants to solve for A and B.

Step 6: General Solution

The general solution to the differential equation y(x) is the sum of the homogeneous and particular solutions: y(x) = y_h(x) + y^∗(x). Insert the constants and fundamental system into this expression.

Step 7: Ansatz for y^{*}(x) for b(x) = cos x + x * e^{-x}

Given b(x) = cos x + x * e^{-x}, propose a particular solution y^{*}(x) of the form y^*(x) = A cos x + B sin x + x(C e^{-x} + D x e^{-x}). This form includes terms that could generate b(x) when differentiated the required number of times.

Key Concepts

homogeneous differential equations

A differential equation is termed as homogeneous if every term is a function of the dependent variable and its derivatives appear only in combination with it, and no separate constant term appears. In the exercise, the given equation is homogeneous:

• \(y^{(5)} + y^{(4)} + y^{(3)} + y^{(2)} = 0\)

The goal is to find the general solution which can be expressed as a linear combination of fundamental solutions. This is done by finding the characteristic equation that simplifies the process of finding these individual solutions.

characteristic equation

To solve the homogeneous differential equation, we use the characteristic equation. We assume the solution to be of the form \(y(x) = e^{rx}\). Substituting this into the homogeneous equation \(y^{(5)} + y^{(4)} + y^{(3)} + y^{(2)} = 0\) yields:

• \(r^5 + r^4 + r^3 + r^2 = 0\)

This polynomial equation is called the characteristic equation, and its roots determine the solutions of the original differential equation. Thus, solving this equation involves factoring and finding its roots, which form the basis for the fundamental solutions of the homogeneous differential equation.

particular solution

For non-homogeneous differential equations, we seek a particular solution that fits the specific form of the non-homogeneous term. In the given problem, we are asked to find a particular solution for \(b(x) = 2 + e^{2x}\). To do this, we can use the method of undetermined coefficients. This involves guessing a form for the particular solution and determining the coefficients by substitution into the original differential equation:

• Assume \(y^*(x) = A + B e^{2x}\)

• Differentiating and substituting back into the equation helps solve for \(A\) and \(B\).

Once found, this particular solution is added to the general solution of the homogeneous equation to form the overall general solution.

method of undetermined coefficients

The method of undetermined coefficients is a systematic procedure used to find a particular solution to non-homogeneous linear differential equations with constant coefficients. It is effective when the non-homogeneous term \(b(x)\) is simple, such as a polynomial, exponential, sine, cosine, or a combination of these. Here's how you typically apply it:

• Guess a form of the particular solution \(y^*(x)\) based on the type of \(b(x)\).
• For example, if \(b(x) = 2 + e^{2x}\), try \(y^*(x) = A + B e^{2x}\).
• Differentiate this guess and substitute it into the differential equation.
• Equate coefficients of like terms to solve for the unknown constants.

Using this method simplifies the process of finding particular solutions by turning the problem into algebraic equations to solve for undetermined coefficients.