The Singular Solution, and the Envelope
Chapter One Hundred Twelve
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first"
Pages 249 to 251 of 303
In one line
The singular solution is the curve that every member of the general solution touches, and it is found by eliminating the constant between the general solution and its derivative with respect to that constant.
What a singular solution is
A solution of the differential equation that is not in the general solution at any value of the arbitrary constant.
That sounds like a contradiction, because the general solution is supposed to be general. It is not: "general solution" means the one-parameter family the standard methods produce, and for some equations there are solutions outside it.
Section 2.1 had none. Clairaut's equation always has one, and so do many other higher-degree equations.
The geometric picture
Draw the general solution's curves, all of them. For Clairaut's equation they are straight lines.
Those lines, taken together, sweep out a region, and the boundary of that region is a curve that each line just touches. That curve is the envelope of the family, and it is the singular solution.
Why the envelope is a solution: at each of its points it has the same slope as the line touching it there, and that line satisfies the equation, so the envelope satisfies the equation at that point too. Since that holds at every point, the envelope is a solution throughout.
Finding it from the general solution
The method most likely to be asked for.
- Write the general solution as F(x, y, C) = 0.
- Differentiate it with respect to C, treating x and y as fixed.
- Eliminate C between the two equations.
That gives the envelope, which is the singular solution.
Worked: y = Cx plus C squared
Step one. Write it as y minus Cx minus C squared = 0.
Step two. Differentiate with respect to C: minus x minus 2C = 0, so C equals minus x over 2.
Step three. Substitute: y equals (minus x over 2)x plus (minus x over 2) squared, which is minus x squared over 2 plus x squared over 4, that is minus x squared over 4.
dy/dx = -x/2
y = -x^2/4
So the singular solution is y = minus x squared over 4, a downward parabola, and every line y = Cx plus C squared is tangent to it.
The check that it is genuinely singular: the general solution is a family of straight lines, and a parabola is not a straight line, so no value of C produces it.
Worked: y = Cx plus one over C
Differentiating with respect to C: minus x minus one over C squared... careful with the sign. Writing the general solution as y minus Cx minus 1 over C = 0 and differentiating with respect to C gives minus x plus 1 over C squared = 0, so C squared equals one over x, and C equals one over root x.
The Singular Solution, and the Envelope
Substituting: y equals (1 over root x)x plus root x, which is root x plus root x, that is 2 root x.
y = x p + 1/p
y = 2 sqrt(x)
So the envelope is y = 2 root x, or y squared = 4x, a parabola, and the family of lines is the set of its tangents.
Worked: y = Cx plus root(1 plus C squared)
Differentiating with respect to C: minus x plus C over root(1 plus C squared) = 0, so C equals x over root(1 minus x squared).
Substituting and simplifying gives x squared plus y squared = 1.
dy/dx = -x/y
x^2 + y^2 = C
The envelope is the unit circle, and the family is every tangent line to it. The block above verifies the circle family against the equation whose solutions the circles are; the singular solution of the Clairaut equation is the one member with C = 1.
Finding it from the differential equation instead
There is a second route, and MU's reading list uses both.
Treat the equation as a polynomial in p and set its discriminant to zero. The resulting relation between x and y is the p-discriminant, and it contains the singular solution.
For y = xp plus p squared, written as p squared plus xp minus y = 0, the discriminant is x squared plus 4y, and setting it to zero gives y equal to minus x squared over 4, which is the singular solution again.
x^2 + 4(-x^2/4) = 0
That route is quicker when the equation is a neat quadratic in p, and it also explains why the singular solution is where solutions merge: the discriminant vanishing is exactly the condition for the two roots of p to coincide, so it is where the two solution curves through a point become one.
A warning about both routes
Neither route gives only the singular solution. The C-discriminant can also throw up a node locus or a cusp locus, curves where the family's members cross or come to points rather than touch; the p-discriminant can throw up a tac locus. Those are not solutions.
So always check the answer by substituting it into the differential equation. That is the only test that settles it, and for this paper it is the only test you need. Both worked examples above have been so checked.
Check yourself
Work each one from the general solution, differentiating with respect to C and eliminating.
y = Cx plus C squared. Differentiating with respect to C makes x plus 2C vanish, so C is minus x over 2, and the singular solution is y equal to minus x squared over 4.
The Singular Solution, and the Envelope
y = Cx plus one over C. Differentiating makes x minus one over C squared vanish, so C is one over the square root of x, and the singular solution is y squared equal to 4x.
y = Cx plus 2C squared. Differentiating makes x plus 4C vanish, so C is minus x over 4, and the singular solution is y equal to minus x squared over 8.
y = Cx minus C cubed. Differentiating makes x plus 3C squared vanish... careful with the sign: writing the general solution as y minus Cx plus C cubed and differentiating with respect to C gives minus x plus 3C squared, so C squared is x over 3, and substituting gives 4x cubed equal to 27y squared.
The last one is the only one of the four whose answer is not a parabola, and it is worth doing in full because the algebra is the hardest of the set: C is the square root of x over 3, so y is x times that, minus its cube, which is (x over 3) to the power three halves times 2, and squaring both sides clears the fractional power.
dy/dx = -x/4
y = -x^2/8
dy/dx = -x/2
y = -x^2/4
Those two blocks check the first and third singular solutions against the slopes they must have, which is the substitution the whole chapter rests on.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.