Question

For the given differential equation, $$ y^{\prime \prime}+4 y=\sin t \cos t+\cos ^{2} 2 t $$

Short answer

Question: Find the general solution of the differential equation \(y'' + 4y = \sin t \cos t + \cos^2 2t\). Answer: The general solution to the given differential equation is \(y(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{2}\sin{t}\cos{t} - \frac{1}{12}\cos^2{2t}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Rewrite the equation in standard form

We have the given differential equation: $$ y'' + 4y = \sin t \cos t + \cos^2 2t $$

Solve the homogeneous equation for the complementary function

The homogeneous equation is given by: $$ y'' + 4y = 0 $$ This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we assume a solution of the form: $$ y = e^{rt} $$ Substituting this into the homogeneous equation, we find the characteristic equation: $$ r^2 + 4 = 0 $$ Solving this quadratic equation, we obtain two complex conjugate roots, \(r_1 = 2i\) and \(r_2 = -2i\). Therefore, the complementary function (CF) is: $$ y_c(t) = C_1 \cos(2t) + C_2 \sin(2t) $$

Guess a particular solution using the undetermined coefficients method

Since we have \(\sin{t}\cos{t}\) and \(\cos^2{2t}\) on the right-hand side of our original equation, we'll guess a particular solution (PS) of the form: $$ y_p(t) = A \sin{t}\cos{t} + B\cos^2{2t} $$ Where \(A\) and \(B\) are constants that we will determine by substituting back into the original equation. First, we calculate the first and second derivatives of \(y_p(t)\): $$ y_p'(t) = A(\cos^2{t} - \sin^2{t}) - 4B\cos{2t}\sin{2t} $$ $$ y_p''(t) = -A(2\sin{t}\cos{t}) - 4B(4\cos^2{2t} - 2) $$ Now, we substitute the particular solution and its second derivative into the original equation: $$ [-A(2\sin{t}\cos{t}) - 4B(4\cos^2{2t} - 2)] + 4[A \sin{t}\cos{t} + B\cos^2{2t}] = \sin{t}\cos{t} + \cos^2{2t} $$ Simplifying and equating coefficients, we get the following system of equations: $$ -2A + 4A = 1 \implies A = \frac{1}{2} $$ $$ -16B + 4B = 1 \implies B = -\frac{1}{12} $$ So, the particular solution is: $$ y_p(t) = \frac{1}{2}\sin{t}\cos{t} - \frac{1}{12}\cos^2{2t} $$

Find the general solution

Finally, we combine the complementary function and the particular solution to obtain the general solution: $$ y(t) = y_c(t) + y_p(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{2}\sin{t}\cos{t} - \frac{1}{12}\cos^2{2t} $$ This is the general solution to the given differential equation.

Key concepts

Homogeneous Differential Equations

When solving differential equations, understanding homogeneous equations is crucial. A homogeneous differential equation is one where every term is a function of the unknown variable and its derivatives. In simpler terms, it's a differential equation that equates to zero when all external sources or inputs are removed, like so:
\[y'' + p(t) y' + q(t) y = 0\]
In our exercise, the homogeneous part is represented by \(y'' + 4y = 0\). To solve this, we typically look for a solution in the exponential form, \(y = e^{rt}\), where \(r\) represents the roots of the characteristic equation, a vital concept we'll discuss later. The solution to the homogeneous equation is known as the complementary function (CF), and it forms one part of the complete solution to the differential equation.

Particular Solution

Differential equations often model systems where external forces or influences are at play, reflected by non-zero terms. Thus, besides the homogeneous equation, we need to find the particular solution (PS) that satisfies the non-homogeneous equation.
A particular solution is a specific solution that includes these non-homogeneous parts. In the initial exercise, we're working with the equation \(y'' + 4y = \text{source terms}\), with source terms being \(\text{sin } t \text{ cos } t + \text{cos}^2 2t\). We guess a form for the PS that resembles these source terms and determine the unknown coefficients by substituting it back into the differential equation. The essence of the PS is to ensure that the combined (CF + PS) equation satisfies the original equation for the non-homogeneous terms as well.

Complementary Function

The complementary function (CF) is an integral component of the general solution to a non-homogeneous differential equation. It represents the solution to the associated homogeneous equation (where the right side is set to zero). For our original problem, the CF corresponds to the solution of the equation \(y'' + 4y = 0\).
By solving the characteristic equation, we derive the CF, which in our case is \(y_c(t) = C_1 \text{ cos}(2t) + C_2 \text{ sin}(2t)\). Here, \(C_1\) and \(C_2\) are arbitrary constants to be determined by boundary conditions or initial values. The CF captures the system's natural behavior without the influence of external forces.

Undetermined Coefficients Method

To find the particular solution (PS) to a non-homogeneous differential equation, we often use the method of undetermined coefficients. This technique involves guessing a solution form based on the non-homogeneous part and then finding the specific coefficients that make the guess satisfy the equation.
The 'guess' is usually structured based on the type of function presented as the non-homogeneous part; common functions include polynomials, exponentials, and trigonometric functions. In our exercise, we guessed the form \(y_p(t) = A \text{ sin}(t)\text{ cos}(t) + B\text{ cos}^2(2t)\). After differentiating and substituting back into the equation, we solved for the coefficients to pinpoint the exact particular solution.

Characteristic Equation

At the heart of solving homogeneous linear differential equations with constant coefficients lies the characteristic equation. For an equation such as \(y'' + p y' + q y = 0\), we assume a solution of the form \(y = e^{rt}\), where \(r\) is to be determined. This leads to a characteristic equation of the form \(r^2 + pr + q = 0\).
The roots of this quadratic equation play a pivotal role in determining the form of the complementary function. In complex root scenarios, like in our example \(r^2 + 4 = 0\), the solutions lead to trigonometric functions in the complementary function. Understanding how to derive and solve the characteristic equation is essential for finding the CF and thus building the complete solution to the homogeneous differential equation.