Chapter 6: Problem 4
Bestimmen Sie die allgemeine Lösung der Differentialgleichungen (a) \(y^{\prime \prime}+2 y^{\prime}+5 y=-\frac{17}{2} \cos 2 x\), (b) \(y^{\prime \prime}-6 y^{\prime}+5 y=4 e^{x}\).
Short Answer
Expert verified
(a) \(y = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) - \frac{17}{2} \cos 2x\), (b) \(y = C_1 e^{5x} + C_2 e^{x} + 2xe^x\).
Step by step solution
01
Solve the Homogeneous Equation for (a)
Consider the homogeneous equation associated with (a): \[ y'' + 2y' + 5y = 0 \]This is a second-order linear homogeneous differential equation with constant coefficients. First, find the characteristic equation, which is obtained by assuming a solution of the form \(y = e^{rx}\) and substituting into the homogeneous equation. This gives us:\[ r^2 + 2r + 5 = 0 \]Solve the characteristic equation using the quadratic formula:\[ r = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{-2 \pm \sqrt{-16}}{2} \]\[ r = -1 \pm 2i \]The roots are complex, so the solution to the homogeneous equation is:\[ y_h = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) \]
02
Find the Particular Solution for (a)
To find the particular solution to the non-homogeneous differential equation \[ y'' + 2y' + 5y = -\frac{17}{2} \cos 2x \]use the method of undetermined coefficients. Assume a particular solution of the form:\[ y_p = A \cos 2x + B \sin 2x \]Substitute \(y_p\) into the left-hand side of the differential equation, and equate coefficients of \( \cos 2x\) and \( \sin 2x\) to solve for \(A\) and \(B\). After substitution, you will find:\[ A = -\frac{17}{2}, \quad B = 0 \]
03
General Solution for (a)
The general solution for equation (a) is the sum of the homogeneous and particular solutions:\[ y = y_h + y_p = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) - \frac{17}{2} \cos 2x \]
04
Solve the Homogeneous Equation for (b)
Consider the homogeneous equation associated with (b):\[ y'' - 6y' + 5y = 0 \]The characteristic equation is\[ r^2 - 6r + 5 = 0 \]Solve it using the quadratic formula:\[ r = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{6 \pm \sqrt{16}}{2} \]\[ r = 5, 1 \]The roots are real and distinct, so the solution to the homogeneous equation is:\[ y_h = C_1 e^{5x} + C_2 e^{x} \]
05
Find the Particular Solution for (b)
The particular solution for \( y'' - 6y' + 5y = 4e^x \) uses the method of undetermined coefficients. Assume a solution of the form:\[ y_p = Axe^x \]Substitute \(y_p\) into the differential equation and solve for \(A\). Upon substitution and simplification:\(A = 2 \), thus:\[ y_p = 2xe^x \]
06
General Solution for (b)
The general solution for equation (b) is the sum of the homogeneous and particular solutions:\[ y = y_h + y_p = C_1 e^{5x} + C_2 e^{x} + 2xe^x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-order Differential Equations
A second-order differential equation is a type of ordinary differential equation (ODE) that involves the second derivative of a function. In this exercise, we dealt with second-order differential equations expressed as \[y'' + ay' + by = f(x)\]where \(y''\) denotes the second derivative, \(ay'\) represents the first derivative term with a coefficient \(a\), \(by\) is the function multiplied by a constant \(b\), and \(f(x)\) is a given function, often referred to as the non-homogeneous part.
This type of equation is important because it arises in various physical contexts, such as oscillations and vibrations, making it a key part of mathematical modeling in physics and engineering.
To solve these equations, you often need to find both a particular solution and a general solution, relying on specific methods suited to the problem.
This type of equation is important because it arises in various physical contexts, such as oscillations and vibrations, making it a key part of mathematical modeling in physics and engineering.
To solve these equations, you often need to find both a particular solution and a general solution, relying on specific methods suited to the problem.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to find a particular solution of a non-homogeneous linear differential equation with constant coefficients. It assumes a particular form for the solution based on the type of the non-homogeneous term \(f(x)\). For example, if \(f(x) = e^{kx}\), a trial solution might be \(y_p = Ae^{kx}\), where \(A\) is an unknown coefficient determined by substituting \(y_p\) into the original equation.
In the exercise, we applied the method to solve equations like:
In the exercise, we applied the method to solve equations like:
- For \(y'' + 2y' + 5y = -\frac{17}{2} \cos 2x\), assume \(y_p = A \cos 2x + B \sin 2x\).
- For \(y'' - 6y' + 5y = 4e^x\), assume \(y_p = Axe^x\).
Characteristic Equation
The characteristic equation is a key tool in solving linear homogeneous differential equations with constant coefficients. It arises from assuming a solution of the form \(y = e^{rx}\) for the homogeneous part and substituting this into the differential equation.
For example, given the equation \[y'' + ay' + by = 0\]you assume a solution \(y = e^{rx}\), leading to the characteristic equation:
For example, given the equation \[y'' + ay' + by = 0\]you assume a solution \(y = e^{rx}\), leading to the characteristic equation:
- \(r^2 + ar + b = 0\)
- If the roots are real and distinct, the solution is of the form \(y_h = C_1 e^{r_1x} + C_2 e^{r_2x}\).
- If the roots are complex, the solution takes the form \(y_h = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x)\).
Homogeneous Solutions
The homogeneous solution of a differential equation involves solving the associated homogeneous equation \(y'' + ay' + by = 0\). This corresponds to the solution of the differential equation when the non-homogeneous part \(f(x) = 0\).
The general form of these solutions depends on the characteristic roots obtained from the characteristic equation:
The general form of these solutions depends on the characteristic roots obtained from the characteristic equation:
- Real and distinct roots produce solutions like \(y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x}\).
- Complex roots result in solutions like \(y_h = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x)\).
- In some cases, even repeated roots lead to solutions that include polynomial terms.