Question

(a) Determine a suitable form for \(Y(t)\) if the method of undetermined coefficients is to be used. (b) Use a computer algebra system to find a particular solution of the given equation. $$ y^{\prime \prime}+4 y=t^{2} \sin 2 t+(6 t+7) \cos 2 t $$

Short answer

Question: Determine the particular solution Yp(t) for the given second-order non-homogeneous linear differential equation: $$ y^{\prime \prime}+4 y=t^{2} \sin 2 t+(6 t+7) \cos 2 t $$ Answer: The particular solution Yp(t) for the given differential equation is: $$ Y_p(t) = (\frac{1}{4}t^2 + \frac{1}{2}t + \frac{1}{4}) \sin(2t) + (\frac{3}{2}t + 2) \cos(2t) $$

Step-by-step solution

Identify the differential equation

We are given the second-order non-homogeneous linear differential equation: $$ y^{\prime \prime}+4 y=t^{2} \sin 2 t+(6 t+7) \cos 2 t $$

Determine a suitable form for Y(t)

We will use the method of undetermined coefficients to determine a suitable form for Y(t). Since the non-homogeneous part contains a polynomial function multiplied by sine and cosine functions, we will assume a solution of the form: $$ Y(t) = (A t^2 + B t + C) \sin(2t) + (D t^2 + E t + F) \cos(2t) $$

Determine the first and second derivative of Y(t)

Now, we will find the first and second derivative of Y(t) with respect to t: First derivative: $$ Y'(t) = 2 (B t + 2 A t - C) \cos(2t) - 2 (E t + 2 D t - F) \sin(2t) $$ Second derivative: $$ Y^{\prime \prime}(t) = - 4 (2 A t + B) \sin(2t) - 4 (2 D t + E) \cos(2t) $$

Substitute Y(t), Y'(t) and Y''(t) into the given differential equation

Now we will substitute Y(t), Y'(t), and Y''(t) into the given differential equation: $$ - 4 (2 A t + B) \sin(2t) - 4 (2 D t + E) \cos(2t) + 4((A t^2 + B t + C) \sin(2t) + (D t^2 + E t + F) \cos(2t)) = t^{2} \sin 2 t+(6 t+7) \cos 2 t $$

Compare the coefficients

Now, we will compare the coefficients for the sine and cosine terms: For the sine terms: $$ - 8 A t -4 B + 4A t^2 + 4B t + 4C = t^2 $$ For the cosine terms: $$ - 8 D t -4 E + 4D t^2 + 4E t + 4F = 6t + 7 $$ We have the following system of equations: 1. \(4A = 1\) 2. \(4B - 8A = 0\) 3. \(4C - 4B = 0\) 4. \(4D = 0\) 5. \(4E - 8D = 6\) 6. \(4F - 4E = 7\)

Solve the system of equations

Solving the system of equations, we get the following values: $$ A = \frac{1}{4} $$ $$ B = \frac{1}{2} $$ $$ C = \frac{1}{4} $$ $$ D = 0 $$ $$ E = \frac{3}{2} $$ $$ F = 2 $$

Write the particular solution Yp(t)

Now we will write the particular solution Yp(t) using the obtained values: $$ Y_p(t) = (\frac{1}{4}t^2 + \frac{1}{2}t + \frac{1}{4}) \sin(2t) + (\frac{3}{2}t + 2) \cos(2t) $$ This is the particular solution of the given second-order non-homogeneous linear differential equation. In an actual problem-solving situation using a computer algebra system, steps 4 to 6 would be performed by the program/framework you use, which would provide you with the coefficients A, B, C, D, E, and F directly.

Key concepts

Second-order Linear Differential Equation

Second-order linear differential equations are pivotal in understanding many physical phenomena, such as oscillation and wave behavior. They involve the second derivative of a function, which makes these equations crucial for modeling systems where acceleration or curvature needs to be described. These equations generally take the standard form: \(a(t)y'' + b(t)y' + c(t)y = g(t)\).
The left-hand side (LHS) of the equation includes terms like the second derivative \(y''\), a term for the first derivative \(y'\), and a term for the original function \(y\). The LHS is sometimes referred to as the homogeneous part of the equation when set to zero.
In the specific problem from the exercise, the differential equation is \(y'' + 4y = t^2 \sin 2t + (6t + 7) \cos 2t\).
Here, \(a(t) = 1\), \(b(t) = 0\), and \(c(t) = 4\), with \(g(t) = t^2 \sin 2t + (6t+7) \cos 2t\), providing the non-homogeneous part. Understanding this allows us to approach solutions using various methods applicable to second-order linear equations.

Non-homogeneous Differential Equation

A non-homogeneous differential equation contains a non-zero term on the right-hand side. This term, usually denoted as \(g(t)\), makes the equation more challenging to solve directly compared to its homogeneous counterpart. Non-homogeneous equations are significant because they allow us to describe systems influenced by external forces or inputs.
The typical structure of a non-homogeneous equation is \(a(t)y'' + b(t)y' + c(t)y = g(t)\), where \(g(t)\) often represents a function that could be a simple polynomial, trigonometric function, exponential, or a combination of these, as seen in the exercise problem.
To solve such equations, we often first find the complementary or homogeneous solution by setting \(g(t) = 0\). Then, we determine the particular solution that directly incorporates the structure of \(g(t)\). Combining both the complementary solution and the particular solution gives the complete response of the system.

Particular Solution

The particular solution is a crucial component used to solve non-homogeneous differential equations. This solution specifically addresses the non-homogeneous part \(g(t)\) of the equation, providing a way to incorporate its influence.
For the exercise's specific equation, \(y'' + 4y = t^2 \sin 2t + (6t+7) \cos 2t\), the particular solution needs to match the form of \(g(t)\). By implementing the method of undetermined coefficients, we propose a form like \(Y(t) = (A t^2 + B t + C) \sin(2t) + (D t^2 + E t + F) \cos(2t)\).
This form mirrors the trigonometric and polynomial nature of \(g(t)\). By substituting this assumed solution form back into the differential equation, you can solve for the unknown coefficients. This is where the method gets its name, as the coefficients are "undetermined" until solved using the equation.
The calculated particular solution for our problem integrates seamlessly with the complementary solution to offer a complete description of the system's behavior.

Computer Algebra System

Computer algebra systems (CAS) are powerful tools for solving complex mathematical problems, including differential equations. These systems can perform symbolic manipulations that automate solving, differentiating, and integrating functions.
Using a CAS, especially for problems involving the method of undetermined coefficients, can significantly simplify the process. Instead of manually calculating derivatives, substituting terms, and solving equations, a CAS handles these steps efficiently and accurately.
In our particular exercise, a CAS can be used to find the particular solution quickly by executing the mathematical steps involved in steps 4 to 6. With the CAS inputting the assumed particular form, extracting derivatives, substituting them into the differential equation, and solving for unknown coefficients become way more efficient. Ultimately, this advantage aids students and professionals alike to focus on understanding the underlying problem rather than cumbersome computations.