00:01
Alright, here we have a differential equation that's drdt of the vector 1 minus 2t and 4t.
00:13
And we want to find the general solution to this differential equation.
00:23
So because we have a derivative here, so we want to actually see what is the original r.
00:32
That could satisfy this differential equation.
00:36
So because this is a derivative, the natural way is to just integrate both sides.
00:44
A trick we can do is to just multiply dt on both sides.
00:48
That gives, and then we integrate both sides.
00:56
All right, so the left side, we're just, integral of the r is just r.
01:00
And the right side, we have the integral of the, d t so that will give inside the vector so that's a t and then we have t squared and then over here we have two t squared all right so this is the the general solution to our differential equation all right so yeah in case you aren't clear of how it that you can see that i just put in the integral and integrated each of these terms individually and with the with the rule that d t and this thing is t this thing here is t squared over 2 all right so we want to find the condition where r of zero i actually i think that we have to remember is that every time we do this into do this integration, we're introducing a constant and one for each dimension.
02:28
So we should plus like plus c and d.
02:35
All right...