When the Two Lines Are Parallel
Chapter Ninety-One
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 201 to 202 of 303
In one line
When the two lines are parallel there is no point to move to, so substitute for the combination they share.
When it happens
The two brackets set to zero are parallel lines exactly when ab' equals a'b, that is when the second bracket's x and y coefficients are a fixed multiple of the first's.
(x + y + 1) dx - (x + y - 1) dy = 0
(2x + 3y - 1) dx - (4x + 6y - 5) dy = 0
In the first, both brackets have coefficients 1 and 1. In the second, the coefficients 4 and 6 are twice 2 and 3. Neither has an intersection to shift to.
Why the previous method fails
Step one of the previous chapter asks you to solve two simultaneous equations. Parallel lines give a pair with no solution: eliminating one unknown eliminates the other as well and leaves a false statement such as 0 = 3.
That is the moment to stop and switch method, and recognising it quickly is the practical point of the ab' minus a'b test.
The substitution
Because the x and y coefficients are proportional, both brackets are functions of the same combination. Put v equal to that combination.
For the first example, both brackets involve x plus y, so put v = x plus y.
For the second, the first bracket is 2x plus 3y minus 1 and the second is 2(2x plus 3y) minus 5, so put v = 2x plus 3y.
Then dv/dx is a constant plus a multiple of dy/dx, so dy/dx can be written in terms of dv/dx, and the whole equation becomes one in v and x, which separates.
Worked
Solve dy/dx = (x plus y plus 1) over (x plus y minus 1).
Put v = x plus y. Then dv/dx equals 1 plus dy/dx, so dy/dx equals dv/dx minus 1.
Substituting: dv/dx minus 1 equals (v plus 1) over (v minus 1).
So dv/dx equals (v plus 1) over (v minus 1) plus 1, which is (v plus 1 plus v minus 1) over (v minus 1), that is 2v over (v minus 1).
Separating: (v minus 1) dv over (2v) equals dx. The left side is one half of (1 minus one over v) dv, which integrates to one half of (v minus log v).
So v minus log v equals 2x plus C, and substituting v = x plus y:
dy/dx = (x + y + 1)/(x + y - 1)
y - x - log(x + y) = C
The x plus y minus log(x plus y) equals 2x plus C rearranges to y minus x minus log(x plus y) equal to a constant, which is the printed form.
When the Two Lines Are Parallel
Worked, with a multiple
Solve dy/dx = (2x plus 3y minus 1) over (4x plus 6y minus 5).
The second bracket is 2(2x plus 3y) minus 5, so put v = 2x plus 3y.
Then dv/dx equals 2 plus 3 dy/dx, so dy/dx equals (dv/dx minus 2) over 3.
Substituting: (dv/dx minus 2) over 3 equals (v minus 1) over (2v minus 5).
So dv/dx equals 2 plus 3(v minus 1) over (2v minus 5), which is (4v minus 10 plus 3v minus 3) over (2v minus 5), that is (7v minus 13) over (2v minus 5).
Separating: (2v minus 5) dv over (7v minus 13) equals dx. The left side is a proper division: two sevenths of (7v minus 13) is 2v minus 26 over 7, so the remainder is minus 9 over 7, and the fraction is two sevenths minus nine sevenths over (7v minus 13).
Integrating gives two sevenths of v, minus nine forty-ninths of the logarithm of (7v minus 13), and that equals x plus a constant.
dy/dx = (2x + 3y - 1)/(4x + 6y - 5)
2(2x + 3y)/7 - 9 log(7(2x + 3y) - 13)/49 - x = C
The exact constants inside the logarithm depend on how the division is arranged, and any equivalent form is correct; what matters is the shape, a linear term in v plus a logarithm of a linear function of v.
The two cases, side by side
| Lines meet | Lines parallel | |
|---|---|---|
| Test | ab' not equal to a'b | ab' equals a'b |
| Substitution | x = X + h, y = Y + k | v = the shared combination |
| What it becomes | homogeneous in X, Y | separable in v and x |
| Then | y = vx | integrate directly |
Both end in an integration; they differ only in how they get there.
Check yourself
dy/dx = (x + y)/(x + y + 1)
(y - x)/2 + log(2x + 2y + 1)/4 = C
For that one, both brackets involve x plus y, so put v = x plus y, giving dv/dx equal to 1 plus v over (v plus 1), which is (2v plus 1) over (v plus 1). Separating gives (v plus 1) dv over (2v plus 1) equal to dx, and dividing out gives one half plus a half over (2v plus 1). Integrating and substituting v = x plus y gives the printed answer.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.