Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{1} & {\sqrt{3}} \\ {\sqrt{3}} & {-1}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{e^{\prime}} \\\ {\sqrt{3} e^{-t}}\end{array}\right) $$

Short answer

The form of the general solution for the given system of differential equations is: $$\mathbf{x}(t) = c_1 e^{\sqrt{2}t}\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right) + c_2 e^{-\sqrt{2}t}\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right) + \left(\begin{array}{c}\frac{2}{\sqrt{3}}e^{t} \\ -\frac{1}{\sqrt{3}}e^{-t}\end{array}\right)$$

Step-by-step solution

Identify the Homogeneous System

The homogeneous system is given by the following matrix equation: $$ \mathbf{x}^{\prime} = \left(\begin{array}{cc}1 & \sqrt{3} \\ \sqrt{3} & -1\end{array}\right)\mathbf{x} $$ We will focus on finding the eigenvalues and eigenvectors for the coefficient matrix to find the homogeneous solution.

Find the Eigenvalues of the Coefficient Matrix

Let A be the given coefficient matrix: $$A = \left(\begin{array}{cc}1 & \sqrt{3} \\ \sqrt{3} & -1\end{array}\right)$$ To find the eigenvalues, solve the characteristic equation given by the determinant of \((A-\lambda I)\): $$|A-\lambda I| = \left|\begin{array}{cc}1-\lambda & \sqrt{3} \\ \sqrt{3} & -1-\lambda\end{array}\right|=(1-\lambda)(-1-\lambda)-(\sqrt{3})^2 = \lambda^2-2$$ Now, we find the roots (eigenvalues) of this equation: \(\lambda^2-2 = 0\), which yields \(\lambda = \pm\sqrt{2}\).

Find the Eigenvectors Corresponding to Each Eigenvalue

Now, we will find the eigenvectors corresponding to each of these eigenvalues by solving the system \((A-\lambda I)\mathbf{v} = 0\). (1) For \(\lambda = \sqrt{2}\): $$\left(\begin{array}{cc}1-\sqrt{2} & \sqrt{3} \\ \sqrt{3} & -1-\sqrt{2}\end{array}\right)\left(\begin{array}{c}v_1 \\ v_2\end{array}\right) = 0$$ We simplify the system and solve for the eigenvector: $$\left(\begin{array}{cc}-1 & \sqrt{3} \\ \sqrt{3} & -3\end{array}\right)\left(\begin{array}{c}v_1 \\ v_2\end{array}\right) = 0$$ \(v_1 = \sqrt{3}v_2 \Rightarrow \mathbf{v}_{\sqrt{2}}=\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right)\) (2) For \(\lambda = -\sqrt{2}\): $$\left(\begin{array}{cc}1+\sqrt{2} & \sqrt{3} \\ \sqrt{3} & -1+\sqrt{2}\end{array}\right)\left(\begin{array}{c}v_1 \\ v_2\end{array}\right) = 0$$ We simplify the system and solve for the eigenvector: $$\left(\begin{array}{cc}3 & \sqrt{3} \\ \sqrt{3} & 1\end{array}\right)\left(\begin{array}{c}v_1 \\ v_2\end{array}\right) = 0$$ \(v_1 = -\frac{1}{\sqrt{3}}v_2 \Rightarrow \mathbf{v}_{-\sqrt{2}}=\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right)\) Thus, the eigenvectors corresponding to the eigenvalues are \(\mathbf{v}_{\sqrt{2}}=\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right)\) and \(\mathbf{v}_{-\sqrt{2}}=\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right)\).

Find the Homogeneous Solution

The homogeneous solution is given by a linear combination of the eigenvectors, each multiplied by an exponential term involving the corresponding eigenvalue: $$\mathbf{x}_{h}(t) = c_1 e^{\sqrt{2}t}\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right) + c_2 e^{-\sqrt{2}t}\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right)$$ Now that we have the homogeneous solution, we will find a particular solution for the given inhomogeneous system.

Find a Particular Solution Using the Method of Undetermined Coefficients

We will guess a particular solution of the form: $$\mathbf{x}_{p}(t) = \left(\begin{array}{c}a_1e^{t} \\ a_2e^{-t}\end{array}\right)$$ By taking the derivative with respect to t, we obtain: $$\mathbf{x}^{\prime}_{p}(t) = \left(\begin{array}{c}a_1e^{t} \\ -a_2e^{-t}\end{array}\right)$$ Substitute the guessed particular solution and its derivative into the given inhomogeneous system: $$\left(\begin{array}{c}a_1e^{t} \\ -a_2e^{-t}\end{array}\right) = \left(\begin{array}{cc}1 & \sqrt{3} \\ \sqrt{3} & -1\end{array}\right)\left(\begin{array}{c}a_1e^{t} \\ a_2e^{-t}\end{array}\right)+\left(\begin{array}{c}e^{t} \\ \sqrt{3}e^{-t}\end{array}\right)$$ Using the matrix multiplication, we have: $$\left(\begin{array}{c}a_1e^{t} \\ -a_2e^{-t}\end{array}\right) = \left(\begin{array}{c}(1+\sqrt{3})a_2e^{-t} + e^{t} \\ (1+\sqrt{3})a_1e^{t} + \sqrt{3}e^{-t}\end{array}\right)$$ Equating the components, we get: $$a_1e^{t} = (1+\sqrt{3})a_2e^{-t} + e^{t} \\ -a_2e^{-t} = (1+\sqrt{3})a_1e^{t} + \sqrt{3}e^{-t}$$ By solving this system of equations, we get \(a_1 = \frac{2}{\sqrt{3}}\) and \(a_2 = -\frac{1}{\sqrt{3}}\). Therefore, our particular solution is: $$\mathbf{x}_{p}(t) = \left(\begin{array}{c}\frac{2}{\sqrt{3}}e^{t} \\ -\frac{1}{\sqrt{3}}e^{-t}\end{array}\right)$$

Combine the Homogeneous and Particular Solutions

Finally, the general solution for the given inhomogeneous system is the sum of the homogeneous and the particular solutions: $$\mathbf{x}(t) = \mathbf{x}_{h}(t) + \mathbf{x}_{p}(t) = c_1 e^{\sqrt{2}t}\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right) + c_2 e^{-\sqrt{2}t}\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right) + \left(\begin{array}{c}\frac{2}{\sqrt{3}}e^{t} \\ -\frac{1}{\sqrt{3}}e^{-t}\end{array}\right)$$ Thus, the general solution of the given system of equations is: $$\mathbf{x}(t) = c_1 e^{\sqrt{2}t}\left(\begin{array}{c}\sqrt{3} \\ 1\end{array}\right) + c_2 e^{-\sqrt{2}t}\left(\begin{array}{c}-1 \\ \sqrt{3}\end{array}\right) + \left(\begin{array}{c}\frac{2}{\sqrt{3}}e^{t} \\ -\frac{1}{\sqrt{3}}e^{-t}\end{array}\right)$$

Key concepts

Eigenvalues and Eigenvectors

When dealing with differential equations, especially systems of equations like the one we have, eigenvalues and eigenvectors play a crucial role in finding solutions. Here’s how they work:

• Eigenvalues: These are special numbers associated with a matrix. For the given matrix \( A \), the eigenvalues are roots from the characteristic equation \(|A-\lambda I| = 0\).
• Eigenvectors: Once we have an eigenvalue, we solve \((A-\lambda I)\mathbf{v} = 0\) to find the corresponding eigenvector \( \mathbf{v} \).

The magic happens when you use these values to express solutions for differential equations. They help capture behaviors like growth or decay in the system. In our example, we calculated \(\lambda = \pm\sqrt{2}\) and found eigenvectors for each. These combinations are key for the homogeneous solution of the system.

Homogeneous Solution

The homogeneous solution of a system is part of the general solution that deals with the associated homogeneous differential equation. It’s characterized by the following:

• It only involves the matrix part \(A\) without any additional terms, like \(\mathbf{x}^\prime = A\mathbf{x}\).
• We use eigenvalues and eigenvectors to write the solution.
• The homogeneous solution reflects natural responses of the system without external inputs.

In this problem, the homogeneous solution is expressed as a linear combination of the eigenvectors, each modified by an exponential function based on their eigenvalues. This allows us to describe all potential behaviors that arise from internal dynamics, before any external inputs are considered.

Particular Solution

To address non-homogeneous parts of a differential system, we need a particular solution. Here’s why it matters:

• The particular solution doesnt' depend on initial conditions, but instead counteracts external influences or forcing functions.
• Various methods exist to determine it, like using undetermined coefficients.

In this problem, we guessed a particular solution form that complements the position of the non-homogeneous term. By substituting it back into the differential equation, we adjusted parameters to balance the equation. This modification accounts for responses directly caused by external prompts represented in the differential system.

Linear Systems

Linear systems of differential equations consist of equations where the unknown function and its derivatives appear linearly. Important features include:

• Superposition: Solutions can be added together thanks to linearity.
• Matrix Representation: Easier to manage multiple equations simultaneously.

The original differential equation given is linear, comprising a matrix and additional terms. Linear algebra techniques, including matrix operations and eigenvalue analysis, greatly simplify solving these systems, offering a clear path to both homogeneous and particular solutions.

Undetermined Coefficients

In differential equations, undetermined coefficients are a technique to find particular solutions. Here's a brief rundown:

• It's a guessing method. You choose a form based on the type of inhomogeneous term (e.g., exponentials, polynomials).
• You substitute this form into the equation and solve for coefficients that satisfy it.

This approach relies on picking a particular solution structure that mirrors the appearance of the non-homogenous term, making it easier to deduce exact coefficients. In our example, we leveraged undetermined coefficients to align with exponential terms, effectively finding a solution that fits the specific external input involved.