Question

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

(a) \( y = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) - \frac{1}{2}\cos 2x \); (b) \( y = C_1 e^x + C_2 e^{5x} + 2x e^x \).

Step-by-step solution

Identify the Type of Differential Equation (a)

The given differential equation is \( y'' + 2y' + 5y = -\frac{17}{2} \cos 2x \). This is a non-homogeneous linear differential equation with constant coefficients. To solve it, we need both the complementary (homogeneous) solution and a particular solution.

Solve the Homogeneous Equation for (a)

The homogeneous counterpart of the equation is \( y'' + 2y' + 5y = 0 \). Find the characteristic equation by assuming \( y = e^{rx} \), which gives us the characteristic equation: \( r^2 + 2r + 5 = 0 \). Solve this using the quadratic formula: \( r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 5}}{2} = -1 \pm 2i \). Hence, the complementary solution is \( y_c = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) \).

Solve for a Particular Solution for (a)

Since the non-homogeneous term is \(-\frac{17}{2} \cos 2x\), we will use the method of undetermined coefficients. Assume a particular solution of the form \( y_p = A\cos 2x + B\sin 2x \). Substitute \( y_p \) and its derivatives into the left-hand side of the differential equation and equate the coefficients of \( \cos 2x \) and \( \sin 2x \) to solve for \( A \) and \( B \). We find that \( A = -\frac{17}{34} = -\frac{1}{2} \), \( B = 0 \). So, the particular solution is \( y_p = -\frac{1}{2}\cos 2x \).

General Solution for (a)

Combine the complementary solution and particular solution: \( y = y_c + y_p = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) - \frac{1}{2}\cos 2x \). This is the general solution for equation (a).

Identify the Type of Differential Equation (b)

The given differential equation is \( y'' - 6y' + 5y = 4e^x \). This is another non-homogeneous linear differential equation with constant coefficients. We will again find the complementary solution and a particular solution.

Solve the Homogeneous Equation for (b)

For the homogeneous equation \( y'' - 6y' + 5y = 0 \), find the characteristic equation assuming \( y = e^{rx} \). This results in \( r^2 - 6r + 5 = 0 \). Solve this quadratic equation to get \( r = 1 \) and \( r = 5 \). Thus, the complementary solution is \( y_c = C_1 e^{x} + C_2 e^{5x} \).

Solve for a Particular Solution for (b)

The non-homogeneous term is \( 4e^x \). Given that \( e^x \) is part of the complementary solution, assume a particular solution of the form \( y_p = Ax e^x \). Substitute this into the left-hand side of the differential equation and solve for \( A \). Through substitution and simplifying, we find \( A = 2 \). Thus, the particular solution is \( y_p = 2x e^x \).

General Solution for (b)

Combine the complementary solution and particular solution: \( y = y_c + y_p = C_1 e^x + C_2 e^{5x} + 2x e^x \). This is the general solution for equation (b).

Key concepts

Non-homogeneous equations

Non-homogeneous differential equations are equations that have a term on the right-hand side that isn't zero. These terms are known as non-homogeneous terms and are often functions of the independent variable, such as sine, cosine, exponential, polynomial, or a combination of these. Unlike homogeneous equations, where the equation is equal to zero, these have an added challenge.
To solve non-homogeneous equations, it's important to calculate two parts: the homogeneous solution, which solves the related homogeneous equation (where the right-hand side is zero), and the particular solution, which accounts for the non-homogeneous part. Together, these solutions combine to form the general solution. This ensures the entire differential equation is satisfied.
Understanding non-homogeneous equations is crucial as they appear frequently in real-world applications, where additional external forces or inputs are present.

Characteristic equation

The characteristic equation is a key concept when solving linear differential equations, especially those with constant coefficients. This equation arises when you solve the homogeneous part of a differential equation.
To find the characteristic equation, substitute a trial solution of the form \( y = e^{rx} \) into the homogeneous differential equation. This substitution leads to a polynomial equation in terms of \( r \), known as the characteristic equation.
For example, for the differential equation \( y'' + 2y' + 5y = 0 \), the characteristic equation is \( r^2 + 2r + 5 = 0 \). Solving this leads to the roots of the equation, which help form the complementary solution. The roots could be real, complex, or repeated, each affecting the solution form differently.

• Real distinct roots yield exponential solutions.
• Complex roots lead to oscillatory (sine and cosine) solutions.
• Repeated roots require solutions including terms raised in powers of \( x \).

Method of undetermined coefficients

The method of undetermined coefficients is a technique for finding a particular solution to certain non-homogeneous linear differential equations. It involves guessing the form of the particular solution based on the non-homogeneous term.
This method is applicable when the non-homogeneous term is a simple function like a polynomial, exponential, sine, or cosine. We assume the particular solution has the same form and modify it slightly if it overlaps with the homogeneous solution.
For example, if your non-homogeneous term is \(-\frac{17}{2} \cos 2x\), start with a guess \( y_p = A \cos 2x + B \sin 2x \). Differentiate \( y_p \) twice, substitute back into the equation, and solve for the unknown coefficients \( A \) and \( B \). For function like \( 4e^x \), if \( e^x \) is part of the homogeneous solution, multiply by \( x \) so in this case use \( Ax e^x \). This iterative method can look tricky but once mastered offers an efficient way to find particular solutions.

Homogeneous solution

The homogeneous solution is the part of the differential equation solution that satisfies the equation when the non-homogeneous term is zero. It's derived from the characteristic equation of the homogeneous differential equation.
For a given differential equation, such as \( y'' + 2y' + 5y = 0 \), solve the characteristic equation \( r^2 + 2r + 5 = 0 \) to find the roots. Each root informs the structure of the homogeneous solution. In this case, the roots are complex \( -1 \pm 2i \), leading to a solution involving exponential decay and oscillation: \( y_c = e^{-x}(C_1 \cos 2x + C_2 \sin 2x) \).
This part of the solution shows behaviors like oscillations, growth, or decay and reflects the natural response of the system described by the differential equation. The constants \( C_1 \) and \( C_2 \) allow the homogeneous solution to adapt to specific initial conditions.

Particular solution

A particular solution addresses the specific non-homogeneous component of a differential equation. It doesn’t contain arbitrary constants because it satisfies the entire non-homogeneous differential equation, not just its homogeneous counterpart.
To determine the particular solution, use the method of undetermined coefficients by assuming a solution form that matches the non-homogeneous part. If the non-homogeneous term is \(-\frac{17}{2} \cos 2x\), the particular solution could be guessed as \( y_p = A \cos 2x + B \sin 2x \). After substituting \( y_p \) and solving, you might find that \( A = -\frac{1}{2} \) and \( B = 0\), giving \( y_p = -\frac{1}{2} \cos 2x \).
When combined with the homogeneous solution, the particular solution forms the general solution to the non-homogeneous differential equation. It's crucial because it ensures that the solution set describes both the natural and the forced responses of the system.