Question

In Exercises \(1-10\) find a particular solution. $$ \mathbf{y}^{\prime}=\frac{1}{5}\left[\begin{array}{rr} -4 & 3 \\ -2 & -11 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} 50 e^{3 t} \\ 10 e^{-3 t} \end{array}\right] $$

Short answer

Question: Find a particular solution for the following linear system of ordinary differential equations with constant coefficients and non-homogeneous terms: $$\mathbf{y}'(t) = \frac{1}{5}\left[\begin{array}{rr}-4 & 3 \\ -2 & -11\end{array}\right]\mathbf{y}(t) + \left[\begin{array}{c}50 e^{3 t} \\ 10 e^{-3 t}\end{array}\right].$$ Answer: The particular solution for the given system of differential equations is: $$\mathbf{y}_p(t) = \left[\begin{array}{c} 10 e^{3t} \\ 20 e^{-3t} \end{array}\right].$$

Step-by-step solution

(Step 1: Guess the particular solution)

We need to guess a form for the particular solution based on the non-homogeneous terms. Notice that they involve exponential functions with arguments \(3t\) and \(-3t\). Thus, let's guess that our particular solution \(\mathbf{y}_p\) takes the form: $$\mathbf{y}_p(t) = \left[\begin{array}{c} A e^{3t} \\ B e^{-3t} \end{array}\right],$$ where \(A\) and \(B\) are undetermined coefficients to be found.

(Step 2: Differentiate the guessed solution)

Compute the derivative of the guessed solution with respect to time \(t\), which we'll denote by \(\mathbf{y}_p'(t)\): $$\mathbf{y}_p'(t) = \left[\begin{array}{c} 3A e^{3t} \\ -3B e^{-3t} \end{array}\right].$$

(Step 3: Plug the guessed solution and its derivative into the system)

Now, we need to plug our guessed solution and its derivative back into the given system of equations: $$\left[\begin{array}{c}3A e^{3t} \\ -3B e^{-3t} \end{array}\right] = \frac{1}{5}\left[\begin{array}{rr}-4 & 3 \\ -2 & -11\end{array}\right]\left[\begin{array}{c}A e^{3t} \\ B e^{-3t} \end{array}\right] + \left[\begin{array}{c}50 e^{3 t} \\ 10 e^{-3 t}\end{array}\right].$$

(Step 4: Solve for the undetermined coefficients)

To find the values of \(A\) and \(B\), let's equate the coefficients of the corresponding exponential terms on both sides of the equation: $$\begin{cases} 3A = \frac{1}{5}(-4A e^{6t} + 3B) + 50e^{3t} \\ -3B = \frac{1}{5}(-2A e^{6t} - 11B) + 10 e^{-3t} \end{cases}.$$ Divide both equations by their respective constants to simplify the equations: $$\begin{cases} A = \frac{1}{15}(-4A e^{6t} + 3B) + \frac{50}{3}e^{3t} \\ B = \frac{1}{15}(2A e^{6t} + 11B) - \frac{10}{3} e^{-3t} \end{cases}.$$ Now, we can solve for \(A\) and \(B\) by equating the coefficients of the exponential terms to zero, since the equations hold for all \(t\): $$\begin{cases} -4A + 3B = \frac{150}{3} & (\text{coeffs. of }e^{3t}) \\ 2A + 11B = - \frac{50}{3} & (\text{coeffs. of }e^{-3t}) \end{cases}.$$ Solving the system, we find: $$A = 10, \quad B = 20.$$

(Step 5: Write down the particular solution)

Finally, we can write down the particular solution \(\mathbf{y}_p(t)\) we found by plugging the values of \(A\) and \(B\) back into our guessed solution: $$\mathbf{y}_p(t) = \left[\begin{array}{c} 10 e^{3t} \\ 20 e^{-3t} \end{array}\right].$$

Key concepts

Non-homogeneous Differential Equations

A differential equation involves derivatives of a function and describes how the function changes. Non-homogeneous differential equations have a special type of term called a "forcing function." This term makes the equation non-homogeneous, as it explicitly involves functions of the independent variable like time \(t\).

In our problem, the forcing function is a vector made up of exponential functions: \( \begin{bmatrix} 50e^{3t} \ 10e^{-3t} \end{bmatrix} \). It distinguishes this system from a homogeneous one, which would have been zero in that position.

Solving a non-homogeneous differential equation often involves finding both the complementary solution (related to the homogeneous part of the equation) and the particular solution (related to the non-homogeneous part).

• The complementary solution deals with the homogeneous equation (where the forcing function is zero).
• The particular solution relates to how the forcing function affects the entire system.

Understanding these two components is key in handling non-homogeneous systems.

Exponential Functions

Exponential functions appear frequently in mathematics and physics because they describe growth and decay processes. An exponential function means that the variable \( t \) is in the exponent, resulting in expressions like \( e^{3t} \) or \( e^{-3t} \).

In our exercise, exponential functions dictate the form for our particular solution. This comes from the fact that the forcing terms are exponential, thus hinting at how solutions might look. For instance, in the guessed solution \( \mathbf{y}_p(t) = \begin{bmatrix} A e^{3t} \ B e^{-3t} \end{bmatrix} \), \( A \) and \( B \) are coefficients.

Exponential functions often arise in differential equations where they represent:

• Natural growth or decay processes, such as population growth or radioactive decay.
• Solutions to linear differential equations with constant coefficients.

These solutions exist because the derivative of an exponential function is proportional to the function itself, making it natural to slip into solutions of differential equations.

System of Equations

A system of equations involves finding solutions that satisfy all equations simultaneously. In the context of differential equations, systems mean we are solving for vectors or groups of unknowns.

In the provided problem, we have a system of linear equations with variables \( A \) and \( B \). These represent unknown coefficients in our guessed particular solution. When equating the two sides of the differential equation, we need to isolate and solve for these unknowns.

Steps involve:

• Equate coefficients of corresponding exponential terms from our derivative and original system.
• Set up simultaneous equations by matching terms separately for each exponential function.
• Solve these simultaneous equations to find \( A \) and \( B \).

Solving a system of equations is crucial, as it allows us to determine specific parameter values that satisfy the original problem. Each solution enables us to construct the particular solution.