This page is a draft and is under active development. A first order system of differential equations that can be written in the form \begin{equation} \label{eq:4.2.1} is
called a \( \textcolor{blue}{\mbox{linear system}} \). The linear system \eqref{eq:4.2.1} can be written in matrix form as \begin{eqnarray*} or more briefly as \begin{equation} \label{eq:4.2.2} where \begin{eqnarray*} We call \(A\) the \( \textcolor{blue}{\mbox{coefficient matrix}} \) of \eqref{eq:4.2.2} and \({\bf f}\) the \( \textcolor{blue}{\mbox{forcing function}} \). We'll say that \(A\) and \({\bf f}\) are \( \textcolor{blue}{\mbox{continuous}} \) if their entries are continuous. If \({\bf f}={\bf 0}\), then \eqref{eq:4.2.2} is \( \textcolor{blue}{\mbox{homogeneous}} \); otherwise, \eqref{eq:4.2.2} is \( \textcolor{blue}{\mbox{nonhomogeneous}} \). An initial value problem for \eqref{eq:4.2.2} consists of finding a solution of \eqref{eq:4.2.2} that equals a given constant vector \begin{eqnarray*} at some initial point \(t_0\). We write this initial value problem as \begin{eqnarray*} The next theorem gives sufficient conditions for the existence of solutions of initial value problems for \eqref{eq:4.2.2}. We omit the proof. Theorem \(\PageIndex{1}\)Suppose the coefficient matrix \(A\) and the forcing function \({\bf f}\) are continuous on \((a,b)\), let \(t_0\) be in \((a,b)\), and let \({\bf k}\) be an arbitrary constant \(n\)-vector. Then the initial value problem \begin{eqnarray*} has a unique solution on \((a,b)\). ProofAdd proof here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example \(\PageIndex{1}\)(a) Write the system \begin{equation} \label{eq:4.2.3} in matrix form and conclude from Theorem \((4.2.1)\) that every initial value problem for \eqref{eq:4.2.3} has a unique solution on \((-\infty,\infty)\). (b) Verify that \begin{equation} \label{eq:4.2.4} is a solution of \eqref{eq:4.2.3} for all values of the constants \(c_1\) and \(c_2\). (c) Find the solution of the initial value problem \begin{equation} \label{eq:4.2.5} (a) The system \eqref{eq:4.2.3} can be written in matrix form as \begin{eqnarray*} An initial value problem for \eqref{eq:4.2.3} can be written as \begin{eqnarray*} Since the coefficient matrix and the forcing function are both continuous on \((-\infty,\infty)\), Theorem \((4.2.1)\) implies that this problem has a unique solution on \((-\infty,\infty)\). (b) If \({\bf y}\) is given by \eqref{eq:4.2.4}, then \begin{eqnarray*} (c) We must choose \(c_1\) and \(c_2\) in \eqref{eq:4.2.4} so that \begin{eqnarray*} which is equivalent to \begin{eqnarray*} Solving this system yields \(c_1=1\), \(c_2=-2\), so \begin{eqnarray*} is the solution of \eqref{eq:4.2.5}. The theory of \(n\times n\) linear systems of differential equations is analogous to the theory of the scalar \(n\)th order equation \begin{equation} \label{eq:4.2.6} as developed in Sections 3.1. For example, by rewriting \eqref{eq:4.2.6} as an equivalent linear system it can be shown that Theorem \((4.2.1)\) implies Theorem \((3.1.1)\) (Exercise \((4.2E.12)\). How do you find the solution to a linear system of differential equations?Solving Differential Equations. Step 1: Use the D notation for the derivative. ... . Step 2: Organize the equations. ... . Step 3: Solve by elimination. ... . Step 4: Solve the differential equation. ... . Step 5: Using elimination, solve for the other variables. ... . Step 6: Using initial conditions, solve for the constants.. How do you find the solution of a differential equation?Here is a step-by-step method for solving them:. Substitute y = uv, and. ... . Factor the parts involving v.. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step). Solve using separation of variables to find u.. Substitute u back into the equation we got at step 2.. What is linear system of differential equation?A system of linear differential equations is a set of linear equations relating a group of functions to their derivatives. Because they involve functions and their derivatives, each of these linear equations is itself a differential equation.
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