Problem 3 - Arbeitsbuch Mathematik für Ingenieure

Question

Bestimmen Sie die allgemeine Lösung der Differentialgleichung $$ y^{\prime \prime \prime}-4 y^{\prime \prime}+5 y^{\prime}-2 y=b(x) $$ mit a) \(b(x)=e^{x}\), b) \(b(x)=\sin 2 x\), c) \(b(x)=e^{x}+\sin 2 x\).

Step-by-Step Solution

Answer

Combine the homogeneous solution with particular solutions for each given \( b(x) \) to find the general solution.

Step 1 - Determine the Homogeneous Solution

Solve the homogeneous equation: \[ y''' - 4y'' + 5y' - 2y = 0 \]. Find the characteristic equation: \[ r^3 - 4r^2 + 5r - 2 = 0 \]. Solve for the roots to find the complementary function.

Step 2 - Solve for Roots of the Characteristic Equation

Using polynomial factoring techniques or the Rational Root Theorem, identify the roots of the characteristic equation. Suppose the roots are complex or repeated and write the general solution of the homogeneous equation.

Step 3 - Particular Solution for b(x) = e^x

For \( b(x) = e^x \), let's assume a particular solution of the form \( y_p = A e^x \). Substitute \( y_p \) and its derivatives into the differential equation and solve for A.

Step 4 - Particular Solution for b(x) = \sin(2x)

For \( b(x) = \sin(2x) \), assume a particular solution of the form \( y_p = B \sin(2x) + C \cos(2x) \). Substitute \( y_p \) and its derivatives into the differential equation and solve for B and C.

Step 5 - Particular Solution for b(x) = e^x + \sin(2x)

For \( b(x) = e^x + \sin(2x) \), use the linearity of differential equations to combine the particular solutions from steps 3 and 4: \( y_p = A e^x + B \sin(2x) + C \cos(2x) \).

Step 6 - General Solution

Add the homogeneous solution to the particular solutions from steps 3, 4, and 5 to obtain the general solution for each case: a) For \( b(x) = e^x \): \( y = y_h + A e^x \).b) For \( b(x) = \sin(2x) \): \( y = y_h + B \sin(2x) + C \cos(2x) \).c) For \( b(x) = e^x + \sin(2x) \): \( y = y_h + A e^x + B \sin(2x) + C \cos(2x) \).

Key Concepts

homogeneous differential equations

A homogeneous differential equation is one in which all the terms involving the dependent variable and its derivatives sum to zero. It can be written as: \[a_n y^{(n)} + a_{n-1} y^{(n-1)} + \text{...} + a_1 y' + a_0 y = 0\] In our exercise, the homogeneous companion of our original differential equation is: \[y''' - 4y'' + 5y' - 2y = 0\] To solve this, we look for solutions of the form \(y = e^{rx} \), leading us to the characteristic equation.

characteristic equation

The characteristic equation is found by substituting \( y = e^{rx} \) into the homogeneous differential equation. For our exercise, this substitution provides us the characteristic equation: \[r^3 - 4r^2 + 5r - 2 = 0\] We solve this polynomial equation to determine the roots, which will indicate the form of the general solution to the homogeneous equation. The roots can be real and distinct, real and repeated, or complex. Each type of root provides different forms for the general solution. For example:

• If you have three distinct real roots, the solution will be a combination of exponentials involving those roots.
• For repeated roots, you will add polynomial terms to avoid duplication of solutions.
• Complex roots contribute sinusoidal (sine and cosine) functions to the solution.

particular solution

To find the particular solution for a non-homogeneous differential equation, we need to guess a form based on the non-homogeneous term on the right-hand side of the equation. The choice of the form depends on the function \(b(x)\):

• For \(b(x) = e^x\), we assume a solution of the form \(y_p = Ae^x\).
• For \(b(x) = \sin(2x)\), assume \(y_p = B \sin(2x) + C \cos(2x)\).
• For a combination such as \(b(x) = e^x + \sin(2x)\), simply add the guessed forms for each part: \(y_p = Ae^x + B \sin(2x) + C \cos(2x)\).

After guessing the form, substitute it into the differential equation and solve for the unknown coefficients (such as \(A\), \(B\), \(C\)). Integrate the particular solution with the homogeneous solution to get the general solution for the differential equation. This step ensures that all potential forms of solution are considered, making the final answer comprehensive.