Question

Find a particular solution. $$ y^{\prime \prime}+3 y^{\prime}+y=(2-6 x) \cos x-9 \sin x $$

Short answer

Question: Find a particular solution to the given second-order nonhomogeneous linear differential equation: $$y'' + 3y' + y = (2-6x)\cos x - 9\sin x$$ Answer: A particular solution to the given equation is $$y_p(x) = (1+3x)\cos x - (8-2x)\sin x$$.

Step-by-step solution

Identify the form of the particular solution

Since the nonhomogeneous term in the given equation is \((2-6x)\cos x - 9\sin x\), we can guess that the particular solution has the form: $$ y_p(x) = (A + Bx)\cos x + (C + Dx)\sin x $$ where A, B, C, and D are constants we need to determine.

Compute the first and second derivatives of the guessed solution

To plug the guessed solution into the given equation, we need to compute its first and second derivatives. Using the product and chain rules, we have: First derivative: $$ y_p'(x) = - (A + Bx)\sin x + B\cos x + (C + Dx)\cos x + D\sin x $$ Second derivative: $$ y_p''(x) = -(A+Bx)\cos x - 2B\sin x +(C+Dx)\sin x - 2D\cos x $$

Plug the derivatives into the given equation

Now, substitute the computed derivatives back into the given equation: $$ (-(A+Bx)\cos x-2B\sin x+(C+Dx)\sin x-2D\cos x) + 3(- (A + Bx)\sin x +B\cos x + (C + Dx)\cos x + D\sin x)+ (A + Bx)\cos x + (C + Dx)\sin x = (2-6 x) \cos x-9 \sin x $$

Separate the coefficients of similar terms

In order to equate the coefficients of similar terms, we can rewrite the obtained equation in the form of matching the coefficients of trigonometric functions: $$ [(1-2D)A + B(1-2D)x]\cos x + [(1-2D)C - 2B + D(1-2D)x]\sin x = (2-6x) \cos x - 9\sin x $$

Set up the system of equations for undetermined coefficients

From the previous step, we obtain the following system of equations for the unknown coefficients A, B, C, and D: $$ 1-2D = 2 \\ (1-2D)B - 6(1-2D) = 0 \\ 1-2D = -9 \\ (1-2D)C - 2B + (-9)(1-2D) = 0 $$

Solve the system of equations

Solving the system: $$ D = \frac{1}{2} \\ A = 2(1-2D) = 1 \\ B = 6(1-2D) = 3 \\ C = \frac{9-2B}{1-2D} = -8 \\ $$

Write the final particular solution

Now that we have all the coefficients, we can write the final particular solution: $$ y_p(x) = (1+3x)\cos x - (8-2x)\sin x $$ This is a particular solution to the given second-order nonhomogeneous linear differential equation.

Key concepts

Particular Solution Method

The particular solution method is an essential concept when solving nonhomogeneous linear differential equations. It involves finding a specific solution that satisfies the nonhomogeneous part of the equation. In the context of our exercise example, the nonhomogeneous part is \[ (2-6x)\cos x - 9\sin x \].

The objective is to propose a function form that 'fits' the uniqueness of the nonhomogeneity. We start by guessing the form of a particular solution \(y_p(x)\) and then determine the unknown coefficients within that form. The form must mirror the types of functions presented in the nonhomogeneous term. If the nonhomogeneous term contains trigonometric functions, exponentials, polynomials, or a combination thereof, the guess similarly reflects these functions with added constants representing unknown coefficients that will be determined later.

In the above exercise, the guess for the particular solution accounts for the \(\cos x\) and \(\sin x\) terms in the form of \[ (A + Bx)\cos x + (C + Dx)\sin x \], where \(A, B, C,\) and \(D\) are placeholders for the coefficients. The derivatives of this proposed solution are then computed and substituted back into the original differential equation, setting the stage for determining the values of the undetermined coefficients through matching coefficients of like terms.

Undetermined Coefficients

The method of undetermined coefficients is a technique used for solving nonhomogeneous linear differential equations where the nonhomogeneous part is a linear combination of functions whose derivatives are also part of this set of functions.Once a particular solution is guessed, we determine the coefficients by taking the derivative or derivatives needed to match the order of the differential equation and then substituting these into the original equation. As seen in our exercise, after finding the derivatives \[ y_p'(x) \] and \[ y_p''(x) \], we insert them into the given equation and match the coefficients for like terms involving \(\cos x\) and \(\sin x\).

Equating Coefficients

We then have a system of linear equations composed of the coefficients of like terms. By equating the coefficients on both sides of the equation, a system is developed. For instance, coefficients of \(\cos x\) on one side of the equation are set equal to those on the other side, and similarly for \(\sin x\) or any other function present. Solving this system reveals the values of the undetermined coefficients.The method of undetermined coefficients is straightforward and mechanical, making it a powerful tool for finding particular solutions quickly if the nonhomogeneous term matches the method’s criteria.

Second-order Differential Equations

Second-order differential equations are dynamic systems characterized by a relation involving the second derivative of an unknown function. They frequently arise in physics and engineering, for instance in modeling spring-mass systems or electrical circuits.The general form of a second-order linear differential equation is given by \[ ay'' + by' + cy = g(x) \], where \(a, b,\) and \(c\) are constants, \(y\) is the unknown function of \(x\), \(y'\) and \(y''\) are the first and second derivatives of \(y\) with respect to \(x\), and \(g(x)\) is a given function called the nonhomogeneous term.

Homogeneous vs. Nonhomogeneous

The equation can be homogeneous (if \(g(x)=0\)) or nonhomogeneous (if \(g(x)eq 0\)). The solution to a second-order nonhomogeneous linear differential equation is the sum of the general solution of the homogeneous equation, which does not involve \(g(x)\), and a particular solution \(y_p\) to the nonhomogeneous equation.In our example, finding \(y_p(x)\) tailored to the specific form of \(g(x)\) represented by \[ (2-6x)\cos x - 9\sin x \] is the core challenge addressed by employing the particular solution method. By defining the structure of \(y_p(x)\), we are working towards the complete solution, which would also involve the homogeneous counterpart had it been part of the specific exercise.