00:01
Okay, so here we got negative 4x minus y minus 1 dx plus x plus y plus 3, dy equals 0.
00:38
And so we are going to use the method of equations with linear coefficients.
00:48
Of the form, if we were to have like a1x, a2y, a3, let's be consistent with the book here, although what you name these really doesn't matter.
01:04
If we would have a1x, b1, y, and c1 here, and then a2x as the coefficient b2y and c3.
01:25
We have an a1 is equal to negative 4, a b1 is equal to negative 1, a c1 is equal to negative 1, x is equal to 1, a2 is equal to 1, b2 is equal to 1, and c3 is equal to 3.
01:41
So then we do a little check.
01:43
A1 times b2 is going to be negative 4 times 1.
01:51
And then i'll do a2, which is 1 times b1, which is negative 1.
01:55
And so these do not match.
02:00
They're not equal to each other.
02:02
So we're going to have to do this tricky little thing where we set x equal to u plus h, y equal to v plus k.
02:28
And then we'll choose h and k, so they satisfy negative 4h minus k minus k minus 1 is equal.
02:37
To 0 and h plus k plus 3 is equal to 0 so these were the linear coefficients um i'll just name these equations 1 and 2 and so if we add 1 2 2 together what we end up getting is negative 3h plus 2 equals 0 so this means that h is going to be equal equals two, two thirds.
03:20
And so we're going to plug this, plug h into equation two.
03:27
And from that, we'll get two thirds plus k plus three is equal to zero.
03:48
Okay, and this tells us that k is equal to negative 11 thirds.
03:59
So we have our h and k sorted out.
04:06
So we set then x being equal to u plus h means that x is equal to u plus two -thirds, and y being equal to v plus k means that y is equal to v minus 11 thirds.
04:25
Dx is just going to be equal to du, and dy is equal to d .v.
04:38
So we're going to plug these values into our original equation, plugging in these values for x and y and d u and d v and what we get is negative 4 which was multiplied by x but now we multiply by u plus two -thirds minus y which is v minus 11 thirds minus 1 d u plus and then we had x so it's going to be u plus two thirds plus y, v minus 11 thirds, plus 3, d .y, but we replace that with dv, equal zero.
05:37
And when you calculate this out, we end up with u plus v, dv, is equal to four u plus v, du, du.
06:07
So i just combined a like terms and then subtracted this first expression to both sides.
06:19
Okay, so now if we do a little bit of division, we would get dv, du is equal to 4u plus v over u plus v, which is the same as 4 plus...