Choosing the Method: A Decision Table for Module 2
Chapter One Hundred Fourteen
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first" and Module 2, "2.1 Equation of the first order and of the first degree"
Pages 254 to 255 of 303
In one line
Read the shape of the equation, not its name, and this table tells you what to do.
Why this chapter exists
You have sixty minutes for four to six answers. The single largest waste of that time is five minutes spent on the wrong method before switching. This table is how to avoid it.
Read it from the shape of the equation. The names of the methods are useless at the moment you need them; the shapes are not.
The decision table for the whole of sections 2.1 and 2.2
| What you see | Method | Chapter |
|---|---|---|
| dy/dx equals a function of x alone | integrate | separation of variables |
| dy/dx equals a product or quotient of a function of x and a function of y | separate the variables | separation of variables |
| dy/dx equals a function of (ax + by + c) | v = ax + by + c | equations that become separable |
| M and N of the same total degree | y = vx | homogeneous equations |
| two linear brackets with constants, lines meeting | shift the origin | non-homogeneous linear |
| two linear brackets with constants, lines parallel | v = the shared combination | the parallel lines case |
| dM/dy equals dN/dx | it is exact; use the shortcut | solving an exact equation |
| dy/dx + Py = Q | the integrating factor e to the integral of P | the linear equation |
| dy/dx + Py = Qy to the n | v = y to the (1 - n) | Bernoulli |
| not linear in y but linear in x | swap the roles | the linear equation |
| p squared or higher, factorises | factorise in p | solvable for p |
| y = xp + a function of p alone | write C for p | Clairaut |
| y easy to isolate | differentiate with respect to x | solvable for y |
| x easy to isolate | differentiate with respect to y | solvable for x |
| y squared or x squared throughout, near Clairaut | substitute for the square | reducing to Clairaut |
| none of the above | look for an integrating factor | the five factor rules |
The order to test in, and how long each test takes
- Is it of first degree in p? One look. If not, go to line 9.
- Does it separate? Ten seconds: try to write it as f(y)dy = g(x)dx.
- Is it exact? Ten seconds: two partial derivatives.
- Is it linear in y? Five seconds: look for y and dy/dx to the first power with no products.
- Is it linear in x? Five seconds: turn it upside down and look again.
- Is it homogeneous? Ten seconds: compare the degrees.
- Is it Bernoulli? Five seconds: a power of y on the right.
- Otherwise hunt for an integrating factor, in the order of that chapter.
- For higher degree: does it factorise in p? Then factorise. Is it Clairaut's form? Then one line. Otherwise solvable for y or for x.
Choosing the Method: A Decision Table for Module 2
Steps one to seven take under a minute between them and settle nearly every question.
The four things that mean you are in the wrong method
An x that will not cancel after putting y = vx. The equation was not homogeneous.
Simultaneous equations with no solution when finding the intersection of two lines. They are parallel; use the other substitution.
An integral you cannot do after separating. Usually the wrong combination was chosen; go back and look for another.
p still in the answer. The elimination step was skipped.
Each of those is a signal rather than a disaster, and noticing it at once is worth more than any amount of algebraic skill.
The checks that cost ten seconds and are always worth it
Substitute the answer back. For an explicit answer, differentiate and substitute. For an implicit one, differentiate the relation implicitly. This is the check the whole of this book has been machine-verified against, and it is the only one that settles the matter.
Count the arbitrary constants. One for a first-order equation, however high its degree; two for a second-order one.
Check a special value. Put x = 0, or y = 0, or whatever is convenient, into both the equation and the answer.
Check the long-term behaviour if the equation came from a model. A cooling body must approach room temperature; a decaying quantity must approach zero.
Worked: reading four equations cold
dy/dx = x y plus x. Factorise the right side as x(y plus 1): separable. Ten seconds.
(2x plus 3y) dx plus (3x plus 2y) dy = 0. Two partial derivatives are both 3: exact. Ten seconds.
x dy/dx plus y = x squared. Divide by x: linear with P equal to one over x. Five seconds.
p squared minus 5p plus 6 = 0. Higher degree, factorises: two families of straight lines. Five seconds.
None of the four took more than ten seconds to classify, and none needed any algebra to do so.
Check yourself
Classify each without solving.
| Equation | Method |
|---|---|
| dy/dx = y/x | separable, or homogeneous |
| dy/dx = (x + y)/(x - y) | homogeneous |
| dy/dx + y tan x = sin x | linear |
| dy/dx + y = y^3 | Bernoulli |
| p^2 - 4 = 0 | factorise in p |
| y = xp + p^2 | Clairaut |
| (y + 2x) dx + (x + 2y) dy = 0 | exact |
The first row has two correct answers, which is normal and is not a problem: take whichever you can do faster.
dy/dx = y/x
y = C x
(y + 2x) dx + (x + 2y) dy = 0
x y + x^2 + y^2 = C
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.