Find the solution of the given initial value problem. Draw the trajectory of the solution in the *x*_{1}*x*_{2}-plane and also draw the graph of *x*_{1} versus *t*.

We can use MATLAB to do the individual calculations, but we need to find the generalized eigenvector 'eta' step-by-step.

>> A = [-5/2 3/2;-3/2 1/2]; x0=[3;2]; % Define A and initial condition vector x0 >> [V,D]=eig(A) V = 0.7071 -0.7071 0.7071 -0.7071 D = -1.0000 0 0 -1.0000 >> L=D(1,1); % L is the repeated eigenvalue. >> v=[V(1,1);V(2,1)]; % v is the repeated eigenvector. >> % 'eye(2)' is the 2x2 identity matrix. >> B=rref([A-L*eye(2) v]) % Solve this system manually to find eta. B = 1.0000 -1.0000 -0.4714 0 0 0 >> eta=[B(1,3);0]; % Found by inspection of B. >> % Solution is X(t)=c1*exp(Lt)*v + c2*exp(Lt)*(tv + eta) >> coeffs=[v eta]\x0; >> c1=coeffs(1); c2=coeffs(2); >> % We now know all pieces of X(t). Use '.*' for multiplication. >> strx1=sprintf('%g*exp(%g*t)*%g + %g*exp(%g*t).*(t*%g+%g)', c1,L,v(1),c2,L,v(1),eta(1)); >> strx2=sprintf('%g*exp(%g*t)*%g + %g*exp(%g*t).*(t*%g+%g)', c1,L,v(2),c2,L,v(2),eta(2)); >> % Convert from strings to functions. >> x1 = inline(strx1,'t'); x2 = inline(strx2,'t'); >> disp(x1) % Here are the solutions. Inline function: x1(t) = 2.82843*exp(-1*t)*0.707107 + -2.12132*exp(-1*t)*(t*0.707107+-0.471405) >> disp(x2) Inline function: x2(t) = 2.82843*exp(-1*t)*0.707107 + -2.12132*exp(-1*t)*(t*0.707107+0) >>

Now we know that the component functions of the solution are

Let's look at a parametric plot of the trajectory in the *x*_{1}*x*_{2}-plane:

>> tpts=-10:0.01:10; % Define t-values >> figure; hold on; >> plot(x1(tpts),x2(tpts)); % Plot parametrically. >> xlabel('x_{1}(t)'); ylabel('x_{2}(t)'); >> title('x_{2}(t) -vs- x_{1}(t)'); >> axis([-.5 2 -.5 2]); % Resize window to show details.

We see deduce that this trajectory is moving toward the origin because the
exponential functions are decaying. Let's look now at a graph of
*x*_{1} -vs- *t* and see how the two graphs relate.

>> figure; hold on; >> plot(tpts,x1(tpts)); % Plot x1(t) -vs- t >> xlabel('t'); ylabel('x_{1}(t)'); title('x_{1}(t) -vs- t'); >> axis([0 10 -.5 2]); % Resize window to see details.

We can estimate the the bottom of the "hook" in the graph of
solution curve occurs around *t* = 3 from looking at the minimum point of
the graph of *x*_{1}. So, as we can see from reading both
graphs together,
*x*_{1} decreases rapidly as *t* goes from negative
infinity to about 3, then *x*_{1} increases again (but quite
slowly) to 0 as *t* goes from about 3 to positive inifinity. We can
intuit the direction of motion from either graph, but only the graph of
*x*_{1} -vs- *t* gives us a sense of how quickly the
motion occurs.