Analyzing Solution to x'=(({{5, -1}, {3, 1}}))x, x(0)=(({{a}, {b}}))

Direction Field: Built from the DE

We can get the direction field without knowing the solution.

((({{(Subscript[x, 1])}, {(Subscript[x, 2])}})))'=(({{5, -1}, {3, 1}}))(({{(Subscript[x, 1])}, {(Subscript[x, 2])}}))=(({{5(Subscript[x, 1])-(Subscript[x, 2])}, {3(Subscript[x, 2])+(Subscript[x, 2])}}))

dx1dt=5x1-x2;
dx2dt=3x1+x2;

field=VectorPlot[{dx1dt,dx2dt},{x1,-30,30},{x2,-30,30},VectorScale->{Tiny,Small,None},FrameLabel->{"(Subscript[x, 1])","(Subscript[x, 2])"},BaseStyle->{20}]

Phase Plot of Solution to IVP Using ParametricPlot

In class we solved the IVP and found

(({{(Subscript[x, 1])}, {(Subscript[x, 2])}}))=(({{((3a-b)/2)Exp[4t]+(1/2)(b-a)Exp[2t]}, {((3a-b)/2)Exp[4t]+(3/2)(b-a)Exp[2t]}}))

x1=((3a-b)/2)Exp[4t]+(1/2)(b-a)Exp[2t];
x2=((3a-b)/2)Exp[4t]+(3/2)(b-a)Exp[2t];

a=-0.64;
b=-2;
ivp=ParametricPlot[{x1,x2},{t,0,2},PlotStyle->{Red,Thick},BaseStyle->20]

Show[field,ivp,FrameLabel->{"(Subscript[x, 1])","(Subscript[x, 2])"},BaseStyle->{20}]

Animation for variety of initial conditions

Clear[a,b]
Manipulate[ParametricPlot[{((3a-b)/2)Exp[4t]+(1/2)(b-a)Exp[2t],((3a-b)/2)Exp[4t]+(3/2)(b-a)Exp[2t]},{t,0,2},PlotStyle->{Red,Thick},PlotRange->{{-30,30},{-30,30}},AxesLabel->{"(Subscript[x, 1])","(Subscript[x, 2])"},BaseStyle->{20}],{a,-2,2},{b,-2,2}]

You Can Use Eliminate to Get Implicit Solution

Clear[a,b,x1,x2]
Eliminate[{x1==(3a-b)/2Exp[4t]+(b-a)/2Exp[2t],x2==(3a-b)/2Exp[4t]+3(b-a)/2Exp[2t]},t]