Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-5} \\ {1} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{-\cos t} \\ {\sin t}\end{array}\right) $$

Short answer

Question: Determine the general solution of the following non-homogeneous system of linear differential equations: $$ \mathbf{x}^{\prime} = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} -\cos t \\ \sin t \end{pmatrix} $$ Answer: The general solution of the given non-homogeneous system of linear differential equations is: $$ \mathbf{x}(t) = \begin{pmatrix} 5e^t & 5e^{-t} \\ e^t & 3e^{-t} \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} + \left(\begin{array}{r}{-\frac{4}{3}\cos t + \frac{2}{3}\sin t} \\\ {-\frac{2}{3}\cos t -\frac{1}{3}\sin t}\end{array}\right) $$ where, \(c_1\) and \(c_2\) are constants, which can be determined depending on the initial conditions of the given problem.

Step-by-step solution

1. Finding the fundamental matrix solution

To find the general solution of the homogeneous system \(\mathbf{x}^{\prime} = A\mathbf{x}\), where \(A=\left(\begin{array}{ll}{2} & {-5} \\\ {1} & {-2}\end{array}\right)\), first, find the eigenvalues and eigenvectors of the matrix A. Solve the following equation for \(\lambda\): $$ \det(A-\lambda I) = \left|\begin{array}{ll}{2-\lambda} & {-5} \\\ {1} & {-2-\lambda} \end{array}\right|= \left(2-\lambda\right)\left(-2-\lambda\right) - (-5)(1), $$ which simplifies to \(\lambda^2=1\), giving eigenvalues \(\lambda_1=1\) and \(\lambda_2=-1\). Now, we'll find the eigenvectors corresponding to these eigenvalues. For \(\lambda_1=1\), solve \((A-\lambda_1I)\mathbf{v}_{1}=0\). We get the following system of equations: $$ \begin{array}{ll}{1} & {-5} \\\ {1} & {-3} \end{array}\mathbf{v}_{1}=0 $$ which gives us the eigenvector \(\mathbf{v}_{1}=\left(\begin{array}{r}{5} \\\ {1}\end{array}\right)\). Similarly, for \(\lambda_2=-1\), solve \((A-\lambda_2I)\mathbf{v}_{2}=0\). We get the following system of equations: $$ \begin{array}{ll}{3} & {-5} \\\ {1} & {-1} \end{array}\mathbf{v}_{2}=0 $$ which gives us the eigenvector \(\mathbf{v}_{2}=\left(\begin{array}{r}{5} \\\ {3}\end{array}\right)\). With these eigenvalues and eigenvectors, we can write the fundamental matrix solution \(\Phi(t)\) as follows: $$ \Phi(t) = \begin{pmatrix} 5e^t & 5e^{-t} \\ e^t & 3e^{-t} \end{pmatrix} $$

2. Finding a particular solution

Now, we need to find a particular solution \(\mathbf{x}_p(t)\) for the non-homogeneous term \(\mathbf{b}(t)=\left(\begin{array}{r}{-\cos t} \\\ {\sin t}\end{array}\right)\). Since we have a sinusoidal function, we'll assume an ansatz particular solution of the form: $$ \mathbf{x}_{p}(t)=\left(\begin{array}{r}{C_{1} \cos t + C_{2} \sin t} \\\ {C_{3} \cos t + C_{4} \sin t}\end{array}\right) $$ Now, let's find the derivative of the ansatz solution: $$ \mathbf{x}_{p}^{\prime}(t)=\left(\begin{array}{r}{-C_{1} \sin t + C_{2} \cos t} \\\ {-C_{3} \sin t + C_{4} \cos t}\end{array}\right) $$ Substitute \(\mathbf{x}_p(t)\) into the given system of differential equations: $$ \left(\begin{array}{r}{-C_{1} \sin t + C_{2} \cos t} \\\ {-C_{3} \sin t + C_{4} \cos t}\end{array}\right) = \left(\begin{array}{ll}{2} & {-5}\\\ {1} & {-2}\end{array}\right) \left(\begin{array}{r}{C_{1} \cos t + C_{2} \sin t} \\\ {C_{3} \cos t + C_{4} \sin t}\end{array}\right) + \left(\begin{array}{r}{-\cos t} \\\ {\sin t}\end{array}\right) $$ This simplifies to: $$ \left(\begin{array}{r}{-C_{1} \sin t + C_{2} \cos t} \\\ {-C_{3} \sin t + C_{4} \cos t}\end{array}\right) = \left(\begin{array}{r}{(2C_{1}-5C_{3})\cos t +(2C_{2}-5C_{4})\sin t - \cos t} \\\ {(C_{1}-2C_{3})\cos t +(C_{2}-2C_{4})\sin t+ \sin t}\end{array}\right) $$ To satisfy this equation, equate the coefficients of the sin(t) and cos(t) terms on both sides. We get the following system of linear equations: $$ \begin{array}{ll} 2C_{1}-5C_{3}=-1 \\ 2C_{2}-5C_{4}=C_{1} \\ C_{1}-2C_{3}=C_{2} \\ C_{2}-2C_{4}=1 \end{array} $$ Solve this linear system to obtain \(C_1=-\frac{4}{3}, C_2=\frac{2}{3}, C_3=-\frac{2}{3},\) and \(C_4=-\frac{1}{3}\). Therefore, \(\mathbf{x}_p(t)=\left(\begin{array}{r}{-\frac{4}{3}\cos t + \frac{2}{3}\sin t} \\\ {-\frac{2}{3}\cos t -\frac{1}{3}\sin t}\end{array}\right)\).

3. Combining solutions to find the general solution

Now, we can write the general solution of the non-homogeneous system by combining the fundamental matrix solution and the particular solution: $$ \mathbf{x}(t) = \Phi(t) \mathbf{c} + \mathbf{x}_p(t) = \begin{pmatrix} 5e^t & 5e^{-t} \\ e^t & 3e^{-t} \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} + \left(\begin{array}{r}{-\frac{4}{3}\cos t + \frac{2}{3}\sin t} \\\ {-\frac{2}{3}\cos t -\frac{1}{3}\sin t}\end{array}\right) $$ where, \(c_1\) and \(c_2\) are constants, which can be determined depending on the initial conditions of the given problem.

Key concepts

Homogeneous Systems

A homogeneous system of linear differential equations is one in which all constant terms are zero. In mathematical terms, a homogeneous system can be represented as \( \mathbf{x}^\prime = A\mathbf{x} \), where \( A \) is a matrix of coefficients and \( \mathbf{x} \) is the vector of unknown functions.

In the context of the exercise, the system without the trigonometric non-homogeneous part \(-\cos t, \sin t\) is the homogeneous part. To solve this, we typically try to find \( \lambda \) (eigenvalues) that satisfy the characteristic equation of the matrix \( A \) and then find the eigenvectors \( \mathbf{v} \) associated with these eigenvalues. The solutions of the homogeneous system are formed by combining these eigenvectors, scaled by \( e^{\lambda t} \).

This procedure is fundamental for solving both homogeneous and non-homogeneous systems of linear differential equations because it provides the complementary solution (also known as the homogeneous solution) to the overall general solution.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in understanding the behavior of linear transformations represented by matrices, and they play a particularly significant role in solving systems of differential equations.

For a square matrix \( A \), an eigenvalue \( \lambda \) is a scalar such that there exists a non-zero vector \( \mathbf{v} \) (the eigenvector) where \( A\mathbf{v} = \lambda \mathbf{v} \). These pairs express how the matrix \( A \) stretches vectors in certain directions. In our exercise, finding the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \) allows us to characterize the solution to the homogeneous system. The corresponding eigenvectors give the directions along which the system evolves over time.

Particular Solution

While the homogeneous solution expresses the behavior of a differential system with no external forces, the particular solution accounts for external inputs or non-zero constant terms in the system. This solution is often denoted as \( \mathbf{x}_p(t) \).

To find a particular solution for the non-homogeneous differential equation given in the exercise, we used an ansatz, a guessed form of solution, usually informed by the nature of the non-homogeneous part. Specifically, for trigonometric non-homogeneous terms like \( -\cos t \) and \( \sin t \), a suitable ansatz is a combination of sine and cosine functions with unknown coefficients. By substituting this ansatz into the differential equation and matching coefficients, we can solve for the unknowns to obtain the particular solution.

Fundamental Matrix Solution

The concept of a fundamental matrix solution is tied to solving a system of linear homogeneous differential equations. It incorporates all the possible solutions to the homogeneous system into a matrix form, called the fundamental matrix \( \Phi(t) \).

The columns of \( \Phi(t) \) are composed of the eigenvectors of the matrix \( A \) scaled by \( e^{\lambda t} \) where \( \lambda \) are the eigenvalues. As seen in the exercise, once \( \Phi(t) \) is constructed, it can be used to solve the homogeneous system by multiplying it with a constant vector \( \mathbf{c} \). The fundamental matrix thus provides a convenient way to express the general solution to the homogeneous system and subsequently, with the inclusion of \( \mathbf{x}_p(t) \) to form the general solution to the non-homogeneous system.