Clairaut's Form
Chapter One Hundred Eleven
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first"
Pages 247 to 248 of 303
In one line
For y = xp plus f(p), the general solution is obtained by writing C in place of p, and nothing else.
MU's label
"Clairaut's form of the equation". It is the easiest method in section 2.2 and the one to look for first, after factorising.
The form
y = x p + f(p)
The x appears only once, multiplied by p, and everything else is a function of p alone. That is the shape to recognise.
| Equation | Clairaut | f(p) |
|---|---|---|
| y = xp + p^2 | yes | p^2 |
| y = xp + 1/p | yes | 1/p |
| y = xp + sqrt(1 + p^2) | yes | sqrt(1 + p^2) |
| y = xp + log p | yes | log p |
| y = x p^2 + p | no | x multiplied by p squared |
| y = 2xp + p^2 | no | the coefficient of x is 2p |
The last two rows are the near misses, and both are real: the fifth has x times p squared and the sixth has a 2 in front. Neither is Clairaut's form as it stands, though the sixth reduces to it, which the chapter on reducing to Clairaut's form shows.
The result
Replace p by C.
y = C x + f(C)
That is the general solution, and it is a family of straight lines. No integration is needed at all.
Why that works
Differentiate y = xp plus f(p) with respect to x.
The left side gives p. The right side, by the product rule, gives p plus x dp/dx plus f prime of p times dp/dx.
So 0 equals (x plus f prime of p) dp/dx, and the equation factorises.
The first factor, dp/dx = 0, says p is a constant. Call it C. Substituting into the original gives y equal to Cx plus f(C), which is the general solution.
The second factor, x plus f prime of p = 0, gives a relation between x and p with no constant in it. Eliminating p between it and the original gives the singular solution, which the next chapter treats.
So Clairaut's equation always has both, and the split is immediate rather than something to hunt for.
Worked: f(p) = p squared
y = x p + p^2
y = C x + C^2
The general solution is the family of straight lines y = Cx plus C squared, one for each C.
And the singular solution, from x plus 2p = 0, so p = minus x over 2, substituted back:
dy/dx = -x/2
y = -x^2/4
Worked: f(p) = one over p
y = x p + 1/p
y = C x + 1/C
The singular solution comes from x minus one over p squared = 0, so p equals one over root x, and substituting gives y equal to root x plus root x, which is 2 root x.
Clairaut's Form
y = x p + 1/p
y = 2 sqrt(x)
Squaring, that is y squared = 4x, a parabola, and every one of the straight lines y = Cx plus 1 over C touches it.
Worked: f(p) = the square root of (1 plus p squared)
This one has a lovely answer and is set regularly.
The general solution is y = Cx plus root(1 plus C squared).
For the singular solution: x plus p over root(1 plus p squared) = 0, so p equals minus x over root(1 minus x squared), and substituting gives, after simplification, x squared plus y squared = 1.
The singular solution is the unit circle, and the general solution is every straight line that touches it. That is the geometry of a tangent line to a circle, and it is why this example appears in every book: the general solution is the set of tangents and the singular solution is the curve they are tangent to.
The procedure, complete
- Check the form: y = xp plus a function of p alone.
- General solution: write C for p. Done.
- For the singular solution: differentiate f, set x plus f prime of p equal to zero, solve for p, and substitute into the original.
- State both.
Step two is one line, which is why this method is worth recognising before trying anything else.
The mistake to avoid
Writing C for p in the wrong equation. The substitution is made in the original equation y = xp plus f(p), not in the differentiated one.
And a second: the singular solution is not obtained by choosing a value of C. It is a separate curve, and no C gives it. A question asking for both wants two distinct answers.
Check yourself
y = x p + p^2
y = C x + C^2
y = x p + 1/p
y = C x + 1/C
y = x p + 2 p^2
y = C x + 2 C^2
y = x p - p^3
y = C x - C^3
All four are the same one-line move. The last one's singular solution comes from x minus 3p squared = 0, so p equals the square root of x over 3, and substituting gives y equal to a multiple of x to the power three halves.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.