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8.1 Preliminary Theory–Linear Systems: Class Notes

8.1a

Parallels exist between concepts, theorems, and vocabulary of linear differential equations and linear systems of differential equations:

• Existence and uniqueness of solutions
• The superposition principle
• Linear dependence/independence and how they relate to the Wronskian (though the Wronskian is different here)
• Fundamental set of solutions
• Homogeneous and nonhomogeneous systems
• Complementary and particular solutions

Recall from algebra: A system of equations can be represented as a matrix equation. For instance, the system

$$\{\begin{array}{c}{\displaystyle 2x+3y-5z=12}\hfill \\ {\displaystyle 7x+4y+6z=18}\hfill \\ {\displaystyle -2x+z=-27}\hfill \end{array}$$

can be represented by the matrix equation

$$\left(\begin{array}{ccc}2& 3& -5\\ 7& 4& 6\\ -2& 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}12\\ 18\\ -27\end{array}\right)$$

Multiplying each row of the coefficient matrix by the column $(x,y,z)$ gives back the left side of each equation, row by row:

$$\left(\begin{array}{c}2x+3y-5z\\ 7x+4y+6z\\ -2x+z\end{array}\right)=\left(\begin{array}{c}12\\ 18\\ -27\end{array}\right)$$

Row 1 (blue) is the first equation, row 2 (green) the second, and row 3 (magenta) the third. The missing $y$ in the third equation becomes a 0 in the matrix.

In this chapter, we will primarily be concerned with linear systems of DEs that have the forms

$$𝐗\prime =𝐀𝐗\text{and}𝐗\prime =𝐀𝐗+𝐅(t)$$

The first form, $𝐗\prime =𝐀𝐗$, is a homogeneous system; the second, with the extra term $𝐅(t)$, is nonhomogeneous.

Example 1

Write the given linear system in matrix form.

$${\displaystyle \{\begin{array}{c}{\displaystyle \frac{dx}{dt}=4x-7y}\hfill \\ {\displaystyle \frac{dy}{dt}=5x}\hfill \end{array}}$$

Solution

The coefficients of $x$ and $y$ fill the rows of the matrix; the second equation has no $y$, so its second entry is 0.

$${\displaystyle \left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right)=\left(\begin{array}{cc}4& -7\\ 5& 0\end{array}\right)\left(\begin{array}{c}x(t)\\ y(t)\end{array}\right)}$$

The left side is $𝐗\prime$, the matrix is $𝐀$, and the column of unknowns is $𝐗$, so this is

Example 2

Write the given linear system in matrix form.

$${\displaystyle \{\begin{array}{c}{\displaystyle \frac{dx}{dt}=-3x+4y+e^{-t}\sin 2t}\hfill \\ {\displaystyle \frac{dy}{dt}=5x+9z+4e^{-t}\cos 2t}\hfill \\ {\displaystyle \frac{dz}{dt}=y+6z-e^{-t}}\hfill \end{array}}$$

Solution

Split each right side into the part with $x$, $y$, $z$ (which goes into $𝐀𝐗$) and the part that depends only on $t$ (which goes into $𝐅(t)$).

$${\displaystyle \left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\\ \frac{dz}{dt}\end{array}\right)=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)+\left(\begin{array}{c}{e}^{-t}\sin 2t\\ 4e^{-t}\cos 2t\\ -e^{-t}\end{array}\right)}$$

This has the form $𝐗\prime =𝐀𝐗+𝐅(t)$. Every entry of $𝐅(t)$ has a factor of $e^{-t}$, so it can be factored out of the column:

Example 3

Write the given linear system without the use of matrices.

$$𝐗\prime =\left(\begin{array}{ccc}7& 5& -9\\ 4& 1& 1\\ 0& -2& 3\end{array}\right)𝐗+\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^{5t}-\left(\begin{array}{c}8\\ 0\\ 3\end{array}\right)e^{-2t}$$

Solution

Here $𝐗=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ and $𝐗\prime =\left(\begin{array}{c}dx/dt\\ dy/dt\\ dz/dt\end{array}\right)$. Row 1 of the matrix times the column $𝐗$ gives the right side of the first equation, and so on; then add the matching entries of the two forcing columns.

8.1b

Example 4

Verify that the vector $𝐗$ is a solution of the given homogeneous linear system.

$$𝐗\prime =\left(\begin{array}{cc}2& 1\\ -1& 0\end{array}\right)𝐗;𝐗=\left(\begin{array}{c}1\\ 3\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)t{e}^t$$

Solution

In components, $x=e^t+4t{e}^t$ and $y=3e^t-4t{e}^t$. Compute both sides of $𝐗\prime =𝐀𝐗$ and compare.

Left side. Differentiate; the product rule gives ${\displaystyle \frac{d}{dt}(t{e}^t)=e^t+t{e}^t}$:

$$\begin{array}{cc}\hfill {\displaystyle 𝐗\prime }& {\displaystyle =\left(\begin{array}{c}1\\ 3\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)(e^t+t{e}^t)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}1\\ 3\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)t{e}^t}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}5\\ -1\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)t{e}^t}\hfill \end{array}$$

Right side. Multiply the matrix by $𝐗$ written as a single column:

$$\begin{array}{cc}\hfill {\displaystyle \left(\begin{array}{cc}2& 1\\ -1& 0\end{array}\right)𝐗}& {\displaystyle =\left(\begin{array}{cc}2& 1\\ -1& 0\end{array}\right)\left(\begin{array}{c}{e}^t+4t{e}^t\\ 3e^t-4t{e}^t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}2e^t+8t{e}^t+3e^t-4t{e}^t\\ -e^t-4t{e}^t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}5e^t+4t{e}^t\\ -e^t-4t{e}^t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}5\\ -1\end{array}\right)e^t+\left(\begin{array}{c}4\\ -4\end{array}\right)t{e}^t}\hfill \end{array}$$

The two sides agree, so $𝐗$ is a solution.