z'=Az with complex eigensystem

Solution derived from eigensystem

Consider z'=Az with the following matrix.  Let's find the eigensystem.

Eigensystem[(({{0, -5}, {1, 0}}))]

Peel off the real and imaginary parts of the eigenvalues and of the eigenvectors.  

(Remember to include the algebraic sign with the real parts but not with the imaginary parts.  This is because we have conjugates.)

Re[{{-I (Sqrt[5]),I (Sqrt[5])},{{-I (Sqrt[5]),1},{I (Sqrt[5]),1}}}]

The above format is {{RealPartEval,RealPartEval},{RealPartEvec,RealPartEvec}}

RealPartEval=0;
RealPartEvec = {0,1};

Im[{{-I (Sqrt[5]),I (Sqrt[5])},{{-I (Sqrt[5]),1},{I (Sqrt[5]),1}}}]

The above format is {{-ImagPartEval,ImagPartEval},{-ImagPartEvec,ImagPartEvec}}.

ImagPartEval =(Sqrt[5]);
ImagPartEvec ={(Sqrt[5]),0};

State the real-valued solutions.

(E^(RealPartEval * t))(RealPartEvec *Cos[ImagPartEval * t] -
ImagPartEvec * Sin[ImagPartEval * t])

(E^(RealPartEval * t))(RealPartEvec *Sin[ImagPartEval * t] +
ImagPartEvec * Cos[ImagPartEval * t])

State the general solution.

Expand[c1 * {-(Sqrt[5]) Sin[(Sqrt[5]) t],Cos[(Sqrt[5]) t]}
+ c2 *{(Sqrt[5]) Cos[(Sqrt[5]) t],Sin[(Sqrt[5]) t]}]

or, solve automatically

Let the vector z have components x and y.  Then Az is...

Clear[x,y]

(({{0, -5}, {1, 0}})).{x[t],y[t]}

Using this form we can use DSolve.

Clear[x,y]
DSolve[{x'[t]==-5 y[t],y'[t]== x[t]},{x[t],y[t]},t]

Notice a slight difference between these results and those of our mechanical method; the latter is a multiple (of 1/2) of the former.  Notice this is fine.

ComplexExpand[(1/2) (E^(-I (Sqrt[5]) t)) (1+(E^(2 I (Sqrt[5]) t))) C[1]+(1/2) I (Sqrt[5]) (E^(-I (Sqrt[5]) t)) (-1+(E^(2 I (Sqrt[5]) t))) C[2]]

TrigExpand[%]

ComplexExpand[-((I (E^(-I (Sqrt[5]) t)) (-1+(E^(2 I (Sqrt[5]) t))) C[1])/(2 (Sqrt[5])))+(1/2) (E^(-I (Sqrt[5]) t)) (1+(E^(2 I (Sqrt[5]) t))) C[2]]

TrigExpand[%]

Define the solutions as functions

xWithCs[t_]:=(Sqrt[5]) C[2] Cos[(Sqrt[5]) t]-(Sqrt[5]) C[1] Sin[(Sqrt[5]) t]
yWithCs[t_]:=C[1] Cos[(Sqrt[5]) t]+C[2] Sin[(Sqrt[5]) t]

Let the initial value of t be zero:  t0=0.

{xWithCs[0],yWithCs[0]}

Look at one solution with initial conditions.

so this one is REALLY easy to solve for the constants.  Let's arbitrarily pick {x0,y0}={4(Sqrt[5]), 3}.

xWithCs[t]/.{C[1]->3,C[2]->4}

Clear[x];
x[t_]:=4 (Sqrt[5]) Cos[(Sqrt[5]) t]-3 (Sqrt[5]) Sin[(Sqrt[5]) t]

yWithCs[t]/.{C[1]->3,C[2]->4}

Clear[y];
y[t_]:=3 Cos[(Sqrt[5]) t]+4 Sin[(Sqrt[5]) t]

Set-up for plots; static plots

Clear[tmin,tmax,xmin,xmax,ymin,ymax]
tmin = 0;  tmax = 9;
xmin = -12.0;  xmax=   12.0;
ymin =-5.5;  ymax=   5.5;

p1 is a static plot of the trajectory of the solution with the stated initial conditions.
p2 and p3 are the component plots.  p2 is shown "sideways" to align its x-axis with the x-axis on the parametric plot.

Clear[p1,p2,p3]
p1=ParametricPlot[{x[t],y[t]},{t,tmin,tmax},AxesLabel->{x,y},PlotRange->{{xmin,xmax},{ymin,ymax}}]

p2=ParametricPlot[{x[t],t},{t,tmin,tmax},AxesLabel->{x,t},PlotRange->{{xmin,xmax},{tmin,tmax}}]

p3=Plot[y[t],{t,tmin,tmax},AxesLabel->{t,y},PlotRange->{{tmin,tmax},{ymin,ymax}}]

A numerical solution and a nice animation.

Clear[numberIntervals,deltat];
numberIntervals=60.0;
deltat=(tmax-tmin)/numberIntervals;

TableForm[Table[N[{t,x[t],y[t]}],{t,tmin,tmax,deltat}]]

Both the table above and the animation below show the direction of the trajectory.

The animation clearly shows the relationship between the component solutions and the parametric solution

Manipulate[GraphicsGrid[{

{Show[   {
ParametricPlot[{x[t],y[t]}, {t,tmin,tmax},
PlotStyle->{Green,Thickness[.015]},PlotRange->{{xmax,xmin},{ymax,ymin}},
AxesLabel->{"x[t]","y[t]"},AspectRatio->Automatic],

Graphics[{       PointSize[Large],Point[{x[q],y[q]}]        }]       }],

Show[    {
ParametricPlot[{t,y[t]},{t,tmin,tmax},
PlotStyle->{Blue,Thickness[.015]},PlotRange->{{tmax,tmin},{ymax,ymin}},
AxesLabel->{"t","y[t]"}],

Graphics[{PointSize[Large],Point[{q,y[q]}] }   ]           }]         },


{Show[    {
ParametricPlot[{x[t],t},{t,tmin,tmax},
PlotStyle->{Red,Thickness[.015]},PlotRange->{{xmax,xmin},{tmax,tmin}},
AxesLabel->{"x[t]","t"}     ],

Graphics[{PointSize[Large],Point[{x[q],q}]      } ]           } ]  }           
} ]       ,
{q,tmin,tmax,deltat}]

Direction Field

Load the Graphics`PlotField` package BEFORE trying to generate the direction field.
Load the package once per Mathematica session.  If you change anything, generate the direction field again.

<< "VectorFieldPlots`"
Clear[p4]
p4 = VectorFieldPlots`VectorFieldPlot[
{-5y1,x1},
{x1, xmin*2,xmax*1.5}, {y1, ymin*1.5,ymax*1.5}];

Generate a list of solutions with different values of the superposition constants.  This represents solutions with different initial conditions.

constList={};
fcnList={};
Do[AppendTo[constList,{c1,c2}];
AppendTo[fcnList,{xWithCs[t],yWithCs[t]}/.{C[1]->c1,C[2]->c2}],
{c1,-5,5,2},{c2,-5,5,2}]
fcnList

Plot all of these solutions.

Clear[p5]
p5=ParametricPlot[Evaluate[fcnList],{t,tmin,tmax}]

Superimpose the solutions on the direction field.

Clear[p6]
p6=Show[{p5,p4},Axes->Automatic,PlotRange->{{xmax 2,xmin 1.5`},{ymax 1.5`,ymin 1.5`}},AxesLabel->{"x[t]","y[t]"},AspectRatio->Automatic]

Describe the system, using appropriate vocabulary.

The origin is a center in the x-y plane.  Centers are stable.

From the eigensystem and our theory, we know that purely imaginary eigenvalues yield centers.

Graph p6 illustrates that the origin is a center.  In particular, the trajectories on p6 are ellipses centered at the origin, and the direction field shows us a counterclockwise rotation around the origin. 

 Both the x and y components of all trajectories oscillate with constant amplitude (though the amplitude differs for each component).  This can be seen in several ways.
*  Consider the specific solution used in p1-p3 and the animation. 
	** p1 shows a center in the x-y plane but doesn't clearly show the direction. We can use the numerical table to indicate a direction.  
	** p2-3 show oscillations in each of the {t,x[t]} and {t,y[t]} planes.
	** The animation shows the correspondence between p1, p2, and p3 as well as the direction.
 * Consider the algebraic representation of the solutions as defined in x[t] and y[t]; the sines and cosines yield oscillations.