Last lesson we saw how to construct general and specific solutions of systems of first order homogeneous differential equations with constant coefficients in the case of real eigenvalues. [br][br]In this lesson we'll consider the case when the eigenvalues are complex. We'll treat everything here without resorting to vector functions. That said, the notation (especially the template for step 4 below) becomes considerably easier if functions are organized in vector form. You can see a similar treatment as this, but using vectorized functions [url=https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/matrix-methods-eigenvalues-and-normal-modes/MIT18_03SCF11_s33_7text.pdf]here[/url].

[br][list=1][*]Be sure the system is in standard form. Be sure you can easily identify the constants [math]x_{11},x_{12},x_{21}[/math] and [math]x_{22}[/math]. [i]Note[/i]: We are using "row first; column second" indexing notation for the elements of the matrix; for example [math]x_{12}[/math] is the coefficient of the entry of the matrix in row 1 and column 2; this is the standard matrix indexing convention in Linear Algebra.[/*][*]In row 1 of the GeoGebra CAS enter the matrix representation of the coefficients as [code]{{x_{11},x_{12}},{x_{21},x_{22}}}[br][/code][/*][*]In row 2, find the eigenvalues of the matrix in row 1 with [code]Eigenvalues($1)[/code]. In these steps we assume these are complex numbers (see the prior lesson for the case of real eigenvectors). We'll only need the first eigenvalue [math]\lambda_1=a+bi[/math]. Be sure to make a note of the real and complex coefficients, [math]a[/math] and [math]b[/math], respectively, of the eigenvalue.[/*][*]In row 3, find the eigenvectors of the matrix in row 1 with [code]Eigenvectors($1)[/code]. Although both columns are eigenvectors, in the case of complex eigenvalues, we will only need the eigenvector in the first column, [math]v_1=\binom{v_{11}+v_{21}i}{v_{12}+v_{22}i}[/math], corresponding the the first eigenvalue [math]\lambda_1[/math] from step 3. Be careful to record the real and complex parts of each entry of this eigenvector.[/*][*]Use the template below to construct the general solutions of the system making use of the real and complex parts of the eigenvalue and the eigenvector.[/*][*]If present, use initial conditions on [math]x[/math] and [math]y[/math] to find the coefficients [code]c_1[/code] and [code]c_2[/code] that yield the specific solution. This will involve solving a system of linear algebraic equations in [code]c_1[/code] and [code]c_2[/code].[/*][*]Check your work by verifying that your results both solve the system of equations [i]and [/i]the match the initial conditions (if present). I strongly recommend using the GeoGebra Algebra Pane. Note that when using GeoGebra, you will need to name the functions something different than [code]x(t)[/code] and [code]y(t)[/code].[/*][/list]Here's the template for step 5:[br][br][math]x\left(t\right)=c_1\cdot e^{a\cdot t}\left(v_{11}\cos\left(b\cdot t\right)-v_{21}\sin\left(b\cdot t\right)\right)+c_2\cdot e^{a\cdot t}\left(v_{11}\sin\left(b\cdot t\right)+v_{21}\cos\left(b\cdot t\right)\right)[/math][br][br][math]y\left(t\right)=c_1\cdot e^{a\cdot t}\left(v_{12}\cos\left(b\cdot t\right)-v_{22}\sin\left(b\cdot t\right)\right)+c_2\cdot e^{a\cdot t}\left(v_{12}\sin\left(b\cdot t\right)+v_{22}\cos\left(b\cdot t\right)\right)[/math]

Let's use this method to solve this system of equations[br][br][math]x'=3x-13y[/math][br][math]y'=5x+1y[/math][br][br]with initial conditions [math]x(0)=-7[/math] and [math]y(0)=5[/math].[br][br][b]1.[/b] [br][br]The coefficients of the system are [math]x_{11}=3,x_{12}=--13,x_{21}=5[/math], and [math]x_{22}=1[/math][br][br][b]2.[/b] and [b]3. [/b]and [b]4. [br][br][/b]We've done steps 2, 3, and 4 in the GeoGebra CAS window below. Note that we only need Eigenvalue 1, and the corresponding first eigenvector (column 1 of the matrix in row 3).

[b]5.[br][br][/b]From the GeoGebra CAS, we can see that the first eigenvalue (which is all we need) is [math]\lambda_1=a+bi=2-8i[/math] and the first eigenvector (which is all we need) is [math]v_1=\binom{v_{11}+v_{21}i}{v_{12}+v_{22}i}=\binom{-3-2i}{1-2i}[/math]. [br][br]Using these pieces of information, we can assemble the solutions with the template. Be careful; it's [i]easy[/i] to get crossed up:[br][br][math]x\left(t\right)=c_1\cdot e^{2t}\left(\left(-3\right)\cdot\cos\left(-8\cdot t\right)-\left(-2\right)\sin\left(-8\cdot t\right)\right)+c_2\cdot e^{2t}\left(\left(-3\right)\sin\left(-8\cdot t\right)+\left(-2\right)\cos\left(-8\cdot t\right)\right)[/math][br][math]y\left(t\right)=c_1\cdot e^{2t}\left(\left(1\right)\cdot\cos\left(-8\cdot t\right)-\left(-2\right)\sin\left(-8\cdot t\right)\right)+c_2\cdot e^{2t}\left(\left(1\right)\sin\left(-8\cdot t\right)+\left(-2\right)\cos\left(-8\cdot t\right)\right)[/math][br][br][b]6.[br][/b][br]The initial conditions we were provided at the outset are [math]x(0)=-7[/math] and [math]y(0)=5[/math]. Plugging 0 in for [math]t[/math] in the general solution, and setting the result equal to these initial conditions yields[br][br][math]x\left(0\right)=c_1\cdot e^{2\cdot0}\left(\left(-3\right)\cdot\cos\left(-8\cdot0\right)-\left(-2\right)\sin\left(-8\cdot0\right)\right)+c_2\cdot e^{2\cdot0}\left(\left(-3\right)\sin\left(-8\cdot0\right)+\left(-2\right)\cos\left(-8\cdot0\right)\right)=-7[/math][br][math]y\left(0\right)=c_1\cdot e^{2\cdot0}\left(\left(1\right)\cdot\cos\left(-8\cdot0\right)-\left(-2\right)\sin\left(-8\cdot0\right)\right)+c_2\cdot e^{2\cdot0}\left(\left(1\right)\sin\left(-8\cdot0\right)+\left(-2\right)\cos\left(-8\cdot0\right)\right)=5[/math][br][br]Recalling that [math]e^0=\cos(0)=1[/math] and [math]\sin(0)=0[/math], this simplifies to:[br][br][br][math]-7=c_1\cdot\left(-3\right)+c_2\cdot\left(-2\right)[/math][br][math]5=c_1\cdot\left(1\right)+c_2\cdot\left(-2\right)[/math][br][br][br]In particular we must solve this system for [code]c_1[/code] and [code]c_2[/code]:[br][br][br][math]-3c_1-2c_2=-7[/math][br][math]c_1-2c_2=5[/math][br][br]This can be done by hand by either substitution, or applying a method from linear algebra. We'll use GeoGebra's CAS however. See below.

[b]7.[/b] [br][br]Let us check to be sure these functions we've produced solve the system of differential equations and the initial conditions. [br][br]We had to call the functions [code]f_x[/code] and [code]f_y [/code]in GeoGebra because [code]x [/code]and [code]y[/code] are not allowable names for functions in GeoGebra.[br][br]Note that--as always--the code for the checks comes from the original system of equations. The code is:[br][br][code]simplify(f_x'-3*f_x+13*f_y)[/code][br][br]and[br][br][code]simplify(f_y'-5*f_x-1*f_y)[br][/code][br]Scroll to the bottom of the Algebra Pane to verify that these both return 0. Also, the initial conditions are checked with [code]f_x(0)[/code] and [code]f_y(0)[/code] both producing the intended values of -7 and 5, respectively.

Try constructing the general solution of the following system which will have purely periodic solutions.[br][br][math]x'=3x+9y[/math][br][math]y'=-4x-3y[/math][br][br]We do not present the worked solution here.[br][br]

It can become confusing switching between phase portraits and the standard view of a function. [br][br]It's worth pausing for a moment to play around with this applet which lets you engage with these two ways of looking at the solutions of a system of first order differential equations. [br][br]On the left Algebra Pane, you can adjust the constant coefficients, as well as the initial conditions. [br][br]On the right are two perspectives of the specific solutions of the differential equation. [br][br][list][*]First, is the phase portrait perspective which shows the specific solution functions, [code]x(t)[/code] and [code]y(t),[/code] plotted as a parametric curve in the slope field of the system of differential equations.[/*][*]Second, is the traditional perspective on the two specific solution functions,  [code]x(t)[/code] and [code]y(t).[/code][/*][/list][br]Try out different sets of coefficients, and different timeframes to get a feel for these two different perspectives.

In the next lesson we'll encounter a system of differential equations that is non linear, and which is used frequently to model species that interact with each other.