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The command should have returned the following result:

prelimsol= NDSolve[{x'[t]== y[t]- x[t]- E^(3t), y'[t]==3 y[t]+2 x[t]-2 E^(-t), x[0]==1,y[0]==1}, {x[t],y[t]},{t,-2,2}]

We've seen this form of solution before, and you may even have been expecting it. One difference, of course, from your previous labs is that the solution actually consists ofÌýtwo parts, the numerical solution forÌýx(t), and the numerical solution forÌýy(t). This makes sense if you think about it.

In the past we have repeatedly used something that I've referred to as the "[[1,1,2]]Ìýtrick." Now that the solution has two parts, we need to be a little more careful. Remember,Ìý[[1,1,2]]Ìýrefers to the result's first row, first column, second part. Looking carefully at our answer, you should be able to see that this refers to theÌýx[t]Ìývalue. So how would we grab theÌýy[t]Ìývalue? Well,Ìýy[t]Ìýis in the first row,ÌýsecondÌýcolumn of the solution, so to extract it's value we'd need to use aÌý[[1,2,2]]Ìýtrick. Be careful not to confuse these two! (If you're having trouble seeing the rows and columns, probably because of the way the matrix wraps around at the edge of the window, feel free to return toÌýMathematicaÌýand redisplay the result using theÌýMatrixFormÌýcommand.)

So let's get down to the job of grabbing the parts of the solution that we wish to use. We're going to pull the expressions forÌýx[t]ÌýandÌýy[t]Ìýinto two separate variables, sayÌýprelimxÌýandÌýprelimyÌýrespectively. We can do so with the following pair of commands:

prelimx=prelimsol[[1,1,2]]
prelimy=prelimsol[[1,2,2]]

Go toÌýMathematica, and issue them both, but make sure toÌýdo them in separate cells! Come back to your browser when you're done.

Now let'sÌýlook atÌýwhat you should have gotten...