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    Home / Differential Equations / Systems of DE's / Solutions to Systems
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    August 27, 2020

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    Section 5-5 : Solutions to Systems

    Now that we’ve got some of the basics out of the way for systems of differential equations it’s time to start thinking about how to solve a system of differential equations. We will start with the homogeneous system written in matrix form,

    \[\begin{equation}\vec x' = A\,\vec x \label{eq:eq1}\end{equation}\]

    where, \(A\) is an \(n \times n\) matrix and \(\vec x\) is a vector whose components are the unknown functions in the system.

    Now, if we start with \(n = 1 \)then the system reduces to a fairly simple linear (or separable) first order differential equation.

    \[x' = ax\]

    and this has the following solution,

    \[x\left( t \right) = c{{\bf{e}}^{a\,t}}\]

    So, let’s use this as a guide and for a general \(n\) let’s see if

    \[\begin{equation}\vec x\left( t \right) = \vec \eta \,{{\bf{e}}^{r\,t}}\label{eq:eq2}\end{equation}\]

    will be a solution. Note that the only real difference here is that we let the constant in front of the exponential be a vector. All we need to do then is plug this into the differential equation and see what we get. First notice that the derivative is,

    \[\vec x'\left( t \right) = r\vec \eta \,{{\bf{e}}^{r\,t}}\]

    So, upon plugging the guess into the differential equation we get,

    \[\begin{align*}r\vec \eta {{\bf{e}}^{r\,t}} & = A\vec \eta {{\bf{e}}^{r\,t}}\\ \left( {A\vec \eta - r\vec \eta } \right){{\bf{e}}^{r\,t}} & = \vec 0\\ \left( {A - rI} \right)\vec \eta {{\bf{e}}^{r\,t}} & = \vec 0\end{align*}\]

    Now, since we know that exponentials are not zero we can drop that portion and we then see that in order for \(\eqref{eq:eq2}\) to be a solution to \(\eqref{eq:eq1}\) then we must have

    \[\left( {A - rI} \right)\vec \eta = \vec 0\]

    Or, in order for \(\eqref{eq:eq2}\) to be a solution to \(\eqref{eq:eq1}\), \(r\) and \(\vec \eta \) must be an eigenvalue and eigenvector for the matrix \(A\).

    Therefore, in order to solve \(\eqref{eq:eq1}\) we first find the eigenvalues and eigenvectors of the matrix \(A\) and then we can form solutions using \(\eqref{eq:eq2}\). There are going to be three cases that we’ll need to look at. The cases are real, distinct eigenvalues, complex eigenvalues and repeated eigenvalues.

    None of this tells us how to completely solve a system of differential equations. We’ll need the following couple of facts to do this.


    1. If \({\vec x_1}\left( t \right)\) and \({\vec x_2}\left( t \right)\) are two solutions to a homogeneous system, \(\eqref{eq:eq1}\), then \[{c_1}{\vec x_1}\left( t \right) + {c_2}{\vec x_2}\left( t \right)\]

      is also a solution to the system.

    2. Suppose that \(A\) is an \(n\) x \(n\) matrix and suppose that \({\vec x_1}\left( t \right)\), \({\vec x_2}\left( t \right)\), …, \({\vec x_n}\left( t \right)\) are solutions to a homogeneous system, \(\eqref{eq:eq1}\). Define, \[X = \left( {{{\vec x}_1}\,\,{{\vec x}_2}\,\, \cdots \,\,{{\vec x}_n}} \right)\]

      In other words, \(X\) is a matrix whose ith column is the \(i^{\text{th}}\) solution. Now define,

      \[W = \det \left( X \right)\]

      We call \(W\) the Wronskian. If \(W \ne 0\) then the solutions form a fundamental set of solutions and the general solution to the system is,

      \[\vec x\left( t \right) = {c_1}{\vec x_1}\left( t \right) + {c_2}{\vec x_2}\left( t \right) + \cdots + {c_n}{\vec x_n}\left( t \right)\]

    Note that if we have a fundamental set of solutions then the solutions are also going to be linearly independent. Likewise, if we have a set of linearly independent solutions then they will also be a fundamental set of solutions since the Wronskian will not be zero.

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