Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{4} & {-2} \\ {8} & {-4}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{t^{-3}} \\\ {-t^{-2}}\end{array}\right), \quad t>0 $$

Short answer

Answer: The general solution to the homogeneous part of the given system is: $$ \mathbf{x}_{h}(t) = c_1 \mathbf{x}_{1} e^{4t} + c_2 \mathbf{x}_{2} e^{-4t} = \left(\begin{array}{c}{c_1 e^{4t} + c_2 e^{-4t}} \\\ {4(c_1 e^{4t} + c_2 e^{-4t})}\end{array}\right) $$ where \(c_1\) and \(c_2\) are constants that depend on the initial conditions.

Step-by-step solution

Solve the homogeneous system

First, we need to find the general solution to the homogeneous part of the given system, which is: $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{4} & {-2} \\\ {8}&{-4}\end{array}\right) \mathbf{x} $$ To do this, we can find the eigenvalues and corresponding eigenvectors of the given matrix. Let's denote the given matrix by A. Then the characteristic equation will be: $$ \det(A - \lambda I) = \left|\begin{array}{cc}{4-\lambda} & {-2} \\\ {8} & {-4-\lambda}\end{array}\right| = (4-\lambda)(-4-\lambda) +(-2)(-8) $$ Solve this equation for \(\lambda\).

Find eigenvalues and eigenvectors

Now, let us solve the characteristic equation to find the eigenvalues: $$ (4-\lambda)(-4-\lambda) +(-2)(-8) = \lambda^2 - 16 $$ The eigenvalues are given by the roots of this equation, i.e., \(\lambda_1 = 4\) and \(\lambda_2 = -4\). Next, we find the eigenvectors corresponding to these eigenvalues. For \(\lambda_1 = 4\): $$ (A - 4I)\mathbf{x}_{1} = \left(\begin{array}{cc}{0} & {-2} \\\ {8} & {-8}\end{array}\right) \left(\begin{array}{c}{x_{1}} \\\ {x_{2}}\end{array}\right) = 0 $$ and for \(\lambda_2 = -4\): $$ (A + 4I)\mathbf{x}_{2} = \left(\begin{array}{cc}{8} & {-2} \\\ {8} & {0}\end{array}\right) \left(\begin{array}{c}{x_{1}} \\\ {x_{2}}\end{array}\right) = 0 $$ Solve these systems to find the eigenvectors, \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\).

Obtain the general homogeneous solution

Upon solving the linear systems, we obtain the eigenvectors: $$ \mathbf{x}_{1} = \left(\begin{array}{c}{1} \\\ {4}\end{array}\right) \text{ and } \mathbf{x}_{2} = \left(\begin{array}{c}{1} \\\ {4}\end{array}\right) $$ The general solution to the homogeneous part of the given system is: $$ \mathbf{x}_{h}(t) = c_1 \mathbf{x}_{1} e^{4t} + c_2 \mathbf{x}_{2} e^{-4t} = \left(\begin{array}{c}{c_1 e^{4t} + c_2 e^{-4t}} \\\ {4(c_1 e^{4t} + c_2 e^{-4t})}\end{array}\right) $$ where \(c_1\) and \(c_2\) are constants that depend on the initial conditions.

Find a particular solution to the given system

Now, we need to find a particular solution to the given inhomogeneous system. Let's guess a solution in the form of: $$ \mathbf{x}_{p}(t) = \left(\begin{array}{c}{a t^{r}} \\\ {b t^{r}}\end{array}\right) $$ Differentiating \(\mathbf{x}_{p}(t)\), we have: $$ \mathbf{x}_{p}^{\prime}(t) = \left(\begin{array}{c}{a r t^{r-1}} \\\ {b r t^{r-1}}\end{array}\right) $$ Plugging \(\mathbf{x}_{p}(t)\) and \(\mathbf{x}_{p}^{\prime}(t)\) into the given system, we can find the values of \(a\), \(b\) and \(r\).

Combine the solutions

Once we have values for \(a\), \(b\), and \(r\), we can write down the particular solution: $$ \mathbf{x}_{p}(t) = \left(\begin{array}{c}{a t^{r}} \\\ {b t^{r}}\end{array}\right) $$ Now, the general solution to the given system will be the sum of the general solution to the homogeneous part and the particular solution: $$ \mathbf{x}(t) = \mathbf{x}_{h}(t) + \mathbf{x}_{p}(t) = \left(\begin{array}{c}{c_1 e^{4t} + c_2 e^{-4t} + a t^{r}} \\\ {4(c_1 e^{4t} + c_2 e^{-4t}) + b t^{r}}\end{array}\right) $$ This is the required general solution to the given system of equations.

Key concepts

Homogeneous System

A homogeneous system in differential equations involves equations where the terms dependent on the function and its derivatives are set to equal a zero vector. This concept is vital because finding solutions to these systems assists in understanding more complex systems comprised of non-homogeneous components. For a system of linear differential equations, it can be written as:\[ \mathbf{x}' = A\mathbf{x} \]where \( A \) is a matrix of coefficients. Solving this system requires examining the matrix to find its eigenvalues and eigenvectors.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a critical role in solving systems of linear differential equations. By finding eigenvalues, we determine crucial characteristics of the matrix associated with the system. The eigenvalues are found from the characteristic equation, derived from the determinant:\[ \det(A - \lambda I) = 0 \]Eigenvectors are the corresponding vectors that transform via the matrix by just scaling with the eigenvalue. For each eigenvalue \( \lambda \), we solve:\[ (A - \lambda I)\mathbf{v} = 0 \]In our exercise, two eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = -4 \) lead to identical eigenvectors, demonstrating dependent solutions normally requiring careful handling via secondary methods or constraints.

Particular Solution

A particular solution to a non-homogeneous system of differential equations is a specific solution that considers external input or forcing functions. In problems containing terms like \( t^{-3} \) or \( -t^{-2} \) seen in our original exercise, we seek a tailored solution format typically guessed by inspection or derived method. The particular solution can take the form:\[ \mathbf{x}_{p}(t) = \begin{pmatrix} a \cdot t^r \ b \cdot t^r \end{pmatrix} \]Plugging this form into the original equation, alongside its derivative, helps find the constants \( a \), \( b \), and exponent \( r \), fine-tuning it for best fit to the inhomogeneous aspects; this addition completes the general solution for the entire system.

General Solution

Constructing the general solution for a system involves combining both the homogeneous and particular solutions. The homogeneous solution emerges from the eigenvectors and eigenvalues of the coefficient matrix. Conversely, the particular solution aligns with the non-zero right-hand side terms of the original equation. Thus, the complete general solution is:\[ \mathbf{x}(t) = \mathbf{x}_{h}(t) + \mathbf{x}_{p}(t) \]This consists, using our specifics, of:

• The homogeneous part taken from eigenvectors: \[ \mathbf{x}_h(t) = c_1 \mathbf{v}_1 e^{4t} + c_2 \mathbf{v}_2 e^{-4t} \]
• The particular solution optimized for the external terms given in the system.

Marrying both parts ensures all aspects of the differential system are fully addressed, addressing initial values and external forces within one cohesive model.