Problem 12 - Advanced Engineering Mathematics

Question

Proceed as in Example 2 to find the general solution of the given differential equation. Use the results obtained in Problems \(1-6 .\) Do not evaluate the integral that defines \(y_{p}(x)\). \(y^{\prime \prime}-2 y^{\prime}+2 y=\cos ^{2} x\)

Step-by-Step Solution

Answer

General solution: \( y(x) = e^x (C_1 \cos x + C_2 \sin x) + y_p(x) \), where \( y_p(x) \) involves an unevaluated integral form.

Step 1: Find the Complementary Function

First, we solve the associated homogeneous equation, which is \( y'' - 2y' + 2y = 0 \). This is a second-order linear differential equation with constant coefficients. The characteristic equation is given by \( r^2 - 2r + 2 = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation \( r^2 - 2r + 2 = 0 \) can be solved using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -2 \), and \( c = 2 \). Thus, \( r = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \).

Step 3: Write the General Solution of the Homogeneous Equation

Since we have complex roots \( 1 \pm i \), the solution to the homogeneous equation is \( y_h(x) = e^x (C_1 \cos x + C_2 \sin x) \), where \( C_1 \) and \( C_2 \) are constants.

Step 4: Use Undetermined Coefficients for Particular Solution

For the non-homogeneous part \( \cos^2 x \), use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \). Therefore, the non-homogeneous term can be addressed using \( y_p(x) = A + B \cos 2x + C \sin 2x \). However, since \( \cos 2x \) and \( \sin 2x \) aren't roots in our homogeneous solution, we can use them directly.

Step 5: Write the General Solution

The general solution is the sum of the complementary function and the particular solution: \( y(x) = y_h(x) + y_p(x) \). Therefore, the general solution of the differential equation is \( y(x) = e^x (C_1 \cos x + C_2 \sin x) + y_p(x) \), where \( y_p(x) = A + B \cos 2x + C \sin 2x \). Set up the integral for \( y_p(x) \) without solving: since undetermined coefficients would lead to integration, express as integral not evaluated yet.

Key Concepts

Second-Order Linear Differential Equations

Second-order linear differential equations are mathematical expressions involving derivatives of a function with respect to one variable, up to the second derivative. These equations have the general form \( a y'' + b y' + c y = f(x) \), where \( a \), \( b \), and \( c \) are constants and \( f(x) \) is a function of \( x \).
These types of equations arise in various areas such as physics, engineering, and finance as they are used to model dynamic systems.

• "Second-order" indicates that the highest derivative is the second derivative \( y'' \).
• "Linear" denotes that the unknown function and its derivatives appear linearly (i.e., not multiplied together or raised to a power other than one).

Furthermore, these differential equations can either be homogeneous, where \( f(x) = 0 \), or non-homogeneous, where \( f(x) \) is not zero.

Characteristic Equation

The characteristic equation helps us solve homogeneous second-order linear differential equations by transforming the problem into an algebraic one. For an equation \( ay'' + by' + cy = 0 \), the characteristic equation is derived by assuming a solution of the form \( y = e^{rx} \).
Plugging \( y = e^{rx} \) into the differential equation gives us the characteristic equation \( ar^2 + br + c = 0 \).
This is a quadratic equation, and its solutions, or "roots," can be found using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The nature of the roots—real and distinct, real and repeated, or complex conjugates—determines the form of the general solution to the differential equation.

Complementary Function

The complementary function, also known as the general solution to the homogeneous equation, is formed using the roots of the characteristic equation. Depending on the nature of these roots, the complementary function is constructed as follows:

• If the roots are real and distinct, \( r_1 \) and \( r_2 \), the complementary function is \( y_h(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \).
• If the roots are real and repeated, \( r \), it becomes \( y_h(x) = (C_1 + C_2 x) e^{rx} \).
• In the case where the roots are complex, \( \alpha \pm i\beta \), the form is \( y_h(x) = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x) \).

The complementary function represents the solution to the equation assuming no external forces are acting on the system (i.e., the case when \( f(x) = 0 \)).

Particular Solution

The particular solution, denoted often as \( y_p(x) \), is a specific solution to the non-homogeneous differential equation, where the function \( f(x) eq 0 \). It accounts for the external forces or input applied to the system.

To find the particular solution, often the method of undetermined coefficients or variation of parameters is used. For the problem in the original exercise, undetermined coefficients are chosen because \( f(x) = \cos^2 x \) is a trigonometric polynomial function.

The steps include assuming a form similar to \( f(x) \) but with undetermined coefficients, then solving for these coefficients by substituting back into the original differential equation and matching terms.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution to a non-homogeneous differential equation. This method works well when \( f(x) \) consists of polynomials, exponentials, or sinusoidal functions—essentially, functions whose derivatives don't produce new forms.

The process entails assuming a form of \( y_p(x) \) that mirrors \( f(x) \) but with coefficients that are "undetermined," meaning they are constants you'll solve for:

• For \( f(x) = \, \cos^2 x \), rewrite it using identities as needed, and assume a trial solution like \( y_p(x) = A + B \cos 2x + C \sin 2x \).
• Substitute \( y_p(x) \) into the differential equation to derive equations for the undetermined coefficients.
• Solve the resulting system of equations to find these coefficients.

The key is matching the assumed solution's form to that of \( f(x) \) to ensure ease of calculation and accuracy.