Question

For each differential equation, (a) Find the complementary solution. (b) Formulate the appropriate form for the particular solution suggested by the method of undetermined coefficients. You need not evaluate the undetermined coefficients. $$ y^{(4)}+8 y^{\prime \prime}+16 y=t \cos 2 t $$

Short answer

The complementary solution is: $$ y_c(t) = C_1 e^{2it} + C_2 e^{-2it} + C_3 e^{(2-2i)t} + C_4 e^{(-2-2i)t} $$ The form of the particular solution is: $$ y_p(t) = (At^2 + Bt + C)\cos{2t} + (Dt^2 + Et + F)\sin{2t} $$

Step-by-step solution

Solve the homogeneous equation

To find the complementary solution, we need to solve the homogeneous equation: $$ y^{(4)} + 8y'' + 16y = 0 $$ Assume the solution to be in the form of \(y=e^{rt}\), where \(r\) is a constant. Substituting this into the homogeneous equation gives: $$ r^4e^{rt} + 8r^2e^{rt} + 16e^{rt} = 0 $$ Factor out \(e^{rt}\): $$ e^{rt}(r^4 + 8r^2 + 16) = 0 $$ Since \(e^{rt}\) never equals zero, we can divide both sides by it, and we're left with a quadratic equation in \(r^2\): $$ r^4 + 8r^2 + 16 = 0 $$ Now, let \(z = r^2\), thus the equation becomes a quadratic equation in \(z\): $$ z^2 + 8z + 16 = 0 $$ By using the quadratic formula or completing the square, we find that the roots are complex, \(z=-4 \pm 4i\). Thus, \(r^2 = -4 \pm 4i\), and the solutions for \(r\) are \(r_1 = 2i\), \(r_2 = -2i\), \(r_3 = 2 - 2i\), and \(r_4 = -2 - 2i\).

Write the complementary solution

Using the solutions for \(r\), we write the complementary solution (also known as the general solution to the homogeneous equation) as: $$ y_c(t) = C_1 e^{2it} + C_2 e^{-2it} + C_3 e^{(2-2i)t} + C_4 e^{(-2-2i)t} $$ where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are constants.

Formulate the form of the particular solution

Now, we need to find the form of the particular solution using the method of undetermined coefficients. Given the right-hand side of the differential equation is $$ t\cos{2t}, $$ we assume the particular solution is in the form of: $$ y_p(t) = (At^2 + Bt + C)\cos{2t} + (Dt^2 + Et + F)\sin{2t} $$ where \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are constants to be determined. We don't need to evaluate these coefficients according to the exercise, so we have our form for the particular solution.

Write the final answer

We have found both the complementary solution and the form of the particular solution. The complementary solution is: $$ y_c(t) = C_1 e^{2it} + C_2 e^{-2it} + C_3 e^{(2-2i)t} + C_4 e^{(-2-2i)t} $$ The form of the particular solution is: $$ y_p(t) = (At^2 + Bt + C)\cos{2t} + (Dt^2 + Et + F)\sin{2t} $$

Key concepts

Complementary Solution

Finding the complementary solution of a differential equation involves solving its associated homogeneous equation. The term 'complementary' refers to this solution being a part of the general solution, which, when added to the particular solution, provides the complete solution of the non-homogeneous equation.

In the given problem, the homogeneous equation is:

• \(y^{(4)} + 8y'' + 16y = 0\)

To solve it, we assume a solution of the form \(y = e^{rt}\), where \(r\) is a constant. We substitute this assumed solution into the equation, leading us to the characteristic equation: \(r^4 + 8r^2 + 16 = 0\).

Solving this characteristic equation using techniques such as factorization or quadratic formulas gives us four roots (in this case complex). The complementary solution is composed of linear combinations of terms involving these roots, and it describes natural oscillations or exponential growth/decay behaviors inherent in the system.

Thus, for this particular problem, the complementary solution is:

• \(y_c(t) = C_1 e^{2it} + C_2 e^{-2it} + C_3 e^{(2-2i)t} + C_4 e^{(-2-2i)t}\)

where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are constants determined by initial conditions.

Homogeneous Equation

A homogeneous equation in the context of differential equations is one where the value on the right side of the equation equals zero. This concept is crucial when determining the complementary solution of any differential equation, as it describes what the system does without external influences or inputs.

In the original exercise, the given differential equation is:

• \(y^{(4)} + 8y'' + 16y = 0\)

This represents our homogeneous equation. Solving it gives the complementary solution, which helps define the natural behavior of the system described by the equation.

The process generally involves assuming a solution of the form \(y = e^{rt}\), then solving for the roots of the resulting characteristic polynomial. The number and nature of these roots (whether real or complex) dictate the form of the complementary solution, capturing oscillatory and/or exponential behavior.

Understanding homogeneous equations helps in isolating the inherent characteristics of a system, separate from any additional forces or inputs described by non-homogeneous parts of the differential equation.

Particular Solution

The particular solution addresses the non-homogeneous part of a differential equation. It specifically counters the non-zero terms of the equation and is essential combined with the complementary solution to form the complete solution.

Using the method of undetermined coefficients, we can assume a form for the particular solution based on the right-hand side of the differential equation. For the given exercise, the equation is:

• \(t \cos 2t\)

To accommodate this, we propose a trial solution with terms that account for both \(\cos 2t\) and \(\sin 2t\), combined with polynomials to match the degree of \(t\) in the equation. This results in:

• \(y_p(t) = (At^2 + Bt + C)\cos 2t + (Dt^2 + Et + F)\sin 2t\)

Here, \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are the coefficients we need to determine in practice, although the exercise does not require their evaluation.

Crafting the particular solution correctly ensures that when substituted back into the differential equation, any external forces or patterns described by the non-homogeneous terms are effectively countered and illustrated in the solution.

Complex Roots

Complex roots arise from the characteristic equation when solving for the complementary solution of a differential equation. They typically take the form of imaginary numbers, marked by a presence of the imaginary unit \(i\), which indicates oscillatory behavior in the solution.

In this example, the characteristic equation derived from the homogeneous equation is:

• \(r^4 + 8r^2 + 16 = 0\)

This results in complex roots because the solutions to this polynomial include terms involving the square root of negative numbers.

The presence of complex roots translates into the complementary solution containing terms such as \(e^{it}\) and \(e^{-it}\), which can be further expressed using Euler's formula into trigonometric functions. This complex behavior embeds sinusoidal components (\(\sin\) and \(\cos\)) to represent oscillations in the system.

Handling complex roots is crucial for systems modeling waves, or oscillations, such as in electrical circuits or mechanical vibratory systems, adding another layer of depth and realism to these mathematical expressions.