Question

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}-y^{\prime}=4+2 \cos 2 t $$

Short answer

Answer: The complementary solution is \(y_c(t) = c_1 + c_2 e^t + c_3 e^{-t}\), the particular solution is \(y_p(t) = 4 + \dfrac{1}{3}\sin{2t}\), and the general solution is \(y(t) = c_1 + c_2 e^t + c_3 e^{-t} + 4 + \dfrac{1}{3}\sin{2t}\).

Step-by-step solution

Find the complementary solution (associated homogeneous equation)

First, we need to find the complementary solution by solving the associated homogeneous equation: $$ y^{\prime \prime \prime} - y^{\prime} = 0. $$ This is a linear homogeneous differential equation. We can solve this by assuming solutions of the form \(y = e^{rt}\). Substituting, we get: $$ r^3 e^{rt} - r e^{rt} = 0. $$ Factoring out \(e^{rt}\), we obtain: $$ e^{rt}(r^3 - r) = 0. $$ Since \(e^{rt}\) is never zero, the characteristic equation is: $$ r^3 - r = 0. $$ Factoring, we get: $$ r(r^2 - 1) = 0. $$ The roots are \(r = 0, \; \pm 1\). Therefore, the complementary solution is given by: $$ y_c(t) = c_1 + c_2 e^t + c_3 e^{-t}. $$

Find a particular solution

Now, we need to find a particular solution to the non-homogeneous equation: $$ y^{\prime \prime \prime} - y^{\prime} = 4 + 2\cos{2t}. $$ For the constant term, \(\textbf{4}\), we use a particular solution of the form: $$ y_p = A. $$ For the term \(\textbf{2cos(2t)}\), we use a particular solution of the form (make sure not to use any form that already appears in the complementary solution): $$ y_p = B\sin{2t} + C\cos{2t}. $$ Thus, combining both particular solutions, we have: $$ y_p = A + B\sin{2t} + C\cos{2t}. $$ Now, we compute the derivatives and substitute them into the given equation: $$ y_p^\prime = 2B\cos{2t} - 2C\sin{2t}, \quad y_p^{\prime\prime} = -4B\sin{2t} - 4C\cos{2t}, \quad y_p^{\prime\prime\prime} = 8B\cos{2t} - 8C\sin{2t}. $$ Substituting, we get: $$ (8B\cos{2t} - 8C\sin{2t}) - (2B\cos{2t} - 2C\sin{2t}) = 4 + 2\cos{2t}. $$ Simplifying and equating coefficients, we have the following system of equations: $$ \begin{cases} 6B = 2 \\ -6C = 0 \\ A = 4. \end{cases} $$ Solving the system, we find \(A = 4, \; B = \dfrac{1}{3}, \; C = 0\). Thus, the particular solution is: $$ y_p(t) = 4 + \dfrac{1}{3}\sin{2t}.

Formulate the general solution

Now, we combine the complementary solution \(y_c(t)\) with the particular solution \(y_p(t)\) to obtain the general solution: $$ y(t) = y_c(t) + y_p(t) = c_1 + c_2 e^t + c_3 e^{-t} + 4 + \dfrac{1}{3}\sin{2t}. $$ The general solution to the given differential equation is: $$ y(t) = c_1 + c_2 e^t + c_3 e^{-t} + 4 + \dfrac{1}{3}\sin{2t}. $$

Key concepts

Complementary Solution

In the world of differential equations, the complementary solution is a vital component. It refers to the solution of the associated homogeneous equation.
An equation is deemed homogeneous when it equals zero, meaning there are no external terms affecting its outcome.
To solve this, we assume a solution form such as \( y = e^{rt} \).

• Substitute this form into the homogeneous equation.
• Rearrange to create a characteristic equation.
• Solve for \( r \) to identify the characteristic roots.

For example, given a differential equation \( y''' - y' = 0 \), we substitute \( y = e^{rt} \), leading to roots \( r = 0, \pm 1 \).
Hence, the complementary solution becomes \( y_c(t) = c_1 + c_2 e^t + c_3 e^{-t} \).
These roots reflect the intrinsic behavior of the system, captured within homogeneous terms.

Particular Solution

A particular solution accounts for the non-homogeneous part of a differential equation.
These equations feature terms that are functions rather than zero.

• Identify each non-homogeneous term.
• Propose a solution form that matches these terms.
• A solution to a constant might take the form \( y_p = A \).
• Solutions for oscillatory terms like \( cos(2t) \) could be \( y_p = B\sin{2t} + C\cos{2t} \).

Substitute the proposed form into the original equation and solve for coefficients.
For \( y''' - y' = 4 + 2\cos{2t} \), the particular solution was derived as \( y_p(t) = 4 + \frac{1}{3}\sin{2t} \).
This addition captures the influence of external factors represented by non-homogeneous terms.

General Solution

The general solution is the complete answer to a differential equation, combining complementary and particular solutions.
In essence, it's the formula that describes all possible behaviors of the system modeled by the differential equation.

• The complementary solution, \( y_c(t) \), appears as the base response.
• The particular solution, \( y_p(t) \), adds the response to external forces.

Together, they construct the general solution \( y(t) = y_c(t) + y_p(t) \).
For our example, the final formulation is \( y(t) = c_1 + c_2 e^t + c_3 e^{-t} + 4 + \frac{1}{3}\sin{2t} \).
This solution encompasses all specific instances of behavior due to both inherent and external influences on the system.

Homogeneous Equation

A homogeneous equation is a type of differential equation where all terms are dependent completely on the function and its derivatives.
There are no external or standalone terms; it looks like \( y^{ extprime extprime extprime} - y^{ extprime} = 0 \).
This makes them simpler as every element constitutes an inherent part of the system itself.

• In solving, convert to the form \( y = e^{rt} \) to identify roots.
• These roots then formulate the complementary solution.

Characteristic equations and their roots are the crux of homogeneous solution finding.
This intrinsic property makes them foundational to deriving complex solutions in broader, non-homogeneous contexts.

Non-Homogeneous Equation

Non-homogeneous equations set themselves apart by their inclusion of free-standing terms.
These terms aren't bound by derivatives of the function and thus represent external inputs into the system.

• An equation such as \( y''' - y' = 4 + 2\cos{2t} \) shows these characteristics.
• Solutions require attention to each individual external function term.

Particular solution methods often involve proposing forms that align with these external terms.
This leads to a specialized solution accommodating influences beyond the internal system dynamics.
Thus, the non-homogeneous component helps capture the full dimension of real-world systems, which seldom operate in isolation.