Equations That Become Separable by a Substitution
Chapter Eighty-Seven
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 191 to 192 of 303
In one line
If the right-hand side depends on x and y only through the combination ax + by + c, substitute for that whole combination.
The shape
dy/dx = f(a x + b y + c)
The right-hand side is a function of one thing: the straight-line combination ax plus by plus c. It is not separable as it stands, because x and y are tangled inside the bracket.
The substitution
Put v equal to the whole combination.
v = a x + b y + c
dv/dx = a + b dy/dx
So dy/dx is (dv/dx minus a) over b, and substituting turns the equation into one in v and x alone, which always separates.
dv/dx = a + b f(v)
Everything on the right depends on v only, so dv over (a plus b f(v)) equals dx, and both sides integrate.
Worked
Solve dy/dx = (x plus y) squared.
Put v = x plus y. Then dv/dx = 1 plus dy/dx, so dy/dx = dv/dx minus 1.
The equation becomes dv/dx minus 1 equals v squared, that is dv/dx equals 1 plus v squared.
Separating: dv over (1 plus v squared) equals dx, so arctan v equals x plus c, and v equals tan(x plus c).
Substituting back: x plus y equals tan(x plus c), so y equals tan(x plus c) minus x.
dy/dx = (x + y)^2
y = tan(x + C) - x
Worked, with coefficients
Solve dy/dx = (2x plus 3y) squared... actually take a simpler one that integrates cleanly.
Solve dy/dx = 2x plus y.
Hmm: that is linear and is better done by the linear method. The substitution method is for when the right-hand side is a function of the combination, not a multiple of it. Take instead:
Solve dy/dx = 1 over (x plus y).
Put v = x plus y, so dy/dx = dv/dx minus 1, and the equation becomes dv/dx minus 1 equals 1 over v, that is dv/dx equals (v plus 1) over v.
Separating: v dv over (v plus 1) equals dx. The left side integrates after dividing out: v over (v plus 1) is 1 minus 1 over (v plus 1), so the integral is v minus log of the size of (v plus 1).
So v minus log(v plus 1) equals x plus c, and substituting back gives the answer.
dy/dx = 1/(x + y)
y - log(x + y + 1) = C
The answer is implicit and cannot be untangled, which is normal for this family.
Worked, one more
Solve dy/dx = cos(x plus y).
Put v = x plus y. Then dv/dx equals 1 plus cos v.
Separating: dv over (1 plus cos v) equals dx. Using the half-angle identity, 1 plus cos v is 2 cos squared of v over 2, so the integral of one half sec squared (v over 2) dv is tan(v over 2).
Equations That Become Separable by a Substitution
So tan((x plus y)/2) equals x plus c.
dy/dx = cos(x + y)
tan((x + y)/2) - x = C
When to recognise it
The signal is that x and y appear only in the same bracket, or only in the same combination, on the right-hand side.
| Equation | Substitute | Why |
|---|---|---|
| dy/dx = (x + y)^2 | v = x + y | the bracket |
| dy/dx = sin(2x - y) | v = 2x - y | inside the sine |
| dy/dx = 1/(x + y + 3) | v = x + y + 3 | the denominator |
| dy/dx = e to the (3x + y) | v = 3x + y | in the exponent |
| dy/dx = x + y | not this method | it is linear, and easier that way |
The last row matters. When the combination appears linearly and on its own, the equation is linear and the linear method is faster. This substitution is for when the combination sits inside something.
The general idea, which recurs
This is the first appearance of what MU calls "Method of substitution" and which she names twice, at 2.1.7 and again at 2.2.6.
The idea is always the same: find the combination of x and y that the equation is really about, name it, and rewrite everything in terms of it. The homogeneous equations of the next chapter use y over x as that combination; Bernoulli's equation uses a power of y; Clairaut's form uses p.
Looking for the right combination is a skill rather than a rule, and the chapter that gathers the substitutions of the whole module into one table is there to help with it.
Check yourself
dy/dx = (x + y + 1)^2
y = tan(x + C) - x - 1
dy/dx = (2x + y)^2
atan((2x + y)/sqrt(2))/sqrt(2) - x = C
The second needs v = 2x plus y, giving dv/dx equal to 2 plus v squared. Separating gives dv over (2 plus v squared) equal to dx, and the standard integral of one over (a squared plus v squared) is one over a times the arctangent of v over a, with a equal to root two here.
That root two is the signature of a constant that is not 1 inside the separated integral, and it is worth expecting rather than suspecting a mistake. Taking the tangent of both sides would give 2x plus y equal to root two times the tangent of root two times (x plus c), which is the explicit form.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.