Equations Homogeneous in x and y
Chapter Eighty-Eight
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 193 to 195 of 303
In one line
If every term of M and N has the same total degree, put y = vx and the equation separates.
MU's label
"Equations homogeneous in x and y".
What homogeneous means here
A function of x and y is homogeneous of degree n if multiplying both x and y by t multiplies the function by t to the n.
In practice: every term has the same total degree, counting the powers of x and y together.
| Function | Homogeneous | Degree |
|---|---|---|
| x^2 + xy + y^2 | yes | 2 |
| x^3 - 2xy^2 | yes | 3 |
| x^2 + y | no | terms of degree 2 and 1 |
| x + y | yes | 1 |
| x^2 + 1 | no | degree 2 and degree 0 |
An equation M dx plus N dy = 0 is homogeneous if M and N are both homogeneous of the same degree.
The word means three different things in this paper
This is worth stopping on, because the confusion is real and it is examined.
Homogeneous in x and y, which is this chapter: every term has the same total degree.
A homogeneous linear equation, which is section 2.3: the right-hand side is zero, as in f(D)y = 0.
Non-homogeneous linear equations, which is MU's 2.1.3: an equation of the form (ax + by + c)dx + (a'x + b'y + c')dy = 0, where the constants c spoil the homogeneity of this chapter's kind.
Three meanings, one word, and the only way through is to read which section you are in.
The test
Divide the equation into the form dy/dx = F(x, y). It is homogeneous exactly when F can be written as a function of y over x alone.
dy/dx = (x^2 + y^2)/(2xy)
Divide the top and the bottom by x squared and the right-hand side becomes (1 plus (y/x) squared) over (2(y/x)), which depends only on y over x. So the equation is homogeneous.
That test is also the reason the substitution works.
Why y = vx always works
Put y = vx, so that v is y over x, and the right-hand side becomes a function of v alone.
Differentiating y = vx by the product rule gives dy/dx equal to v plus x dv/dx.
So the equation becomes v plus x dv/dx equal to F(v), that is:
x dv/dx = F(v) - v
and that always separates: dv over (F(v) minus v) equals dx over x.
That is the whole method, and it works for every homogeneous equation without exception.
The procedure
- Check that the equation is homogeneous, by degrees or by the y over x test.
- Put y = vx and dy/dx = v plus x dv/dx.
- Simplify until the v terms are on one side and the x terms on the other.
- Integrate. The right side is always dx over x, giving a logarithm of x.
- Put v = y over x back in.
Equations Homogeneous in x and y
Step four is worth knowing in advance: the x side of the integration is always log x, so the answer always contains a logarithm of x, and the constant is usually best written as log C so that the logarithms combine.
Worked
Solve dy/dx = (x plus y) over x.
Homogeneous: both top and bottom are of degree 1. Putting y = vx gives v plus x dv/dx equal to (x plus vx) over x, which is 1 plus v.
So x dv/dx equals 1, and dv equals dx over x. Integrating gives v equal to log x plus c, and substituting back gives y over x equal to log x plus c.
dy/dx = (x + y)/x
y = x log(x) + C x
Worked, the classic one
Solve dy/dx = (x squared plus y squared) over (2xy).
Putting y = vx: v plus x dv/dx equals (1 plus v squared) over (2v).
So x dv/dx equals (1 plus v squared) over (2v) minus v, which is (1 minus v squared) over (2v).
Separating: 2v dv over (1 minus v squared) equals dx over x. The left side integrates to minus the logarithm of the size of (1 minus v squared).
So minus log(1 minus v squared) equals log x plus c, and exponentiating gives 1 minus v squared equal to C over x.
Substituting v = y over x and multiplying by x squared: x squared minus y squared equals Cx.
dy/dx = (x^2 + y^2)/(2xy)
(x^2 - y^2)/x = C
The trap in step three
After substituting, every x must cancel. If an x is left over after simplifying, either the equation was not homogeneous or the algebra has gone wrong.
That cancellation is the whole point: it is what leaves an equation in v and x that separates. Checking that it has happened is a free test on step two.
Check yourself
dy/dx = y/x
y = C x
dy/dx = (x + y)/(x - y)
atan(y/x) - log(x^2 + y^2)/2 = C
The second is the standard homogeneous question and it is worth working in full. Putting y = vx gives x dv/dx equal to (1 plus v squared) over (1 minus v). Separating gives (1 minus v) dv over (1 plus v squared) equal to dx over x, and the left side splits into two standard integrals: one over (1 plus v squared), which gives an arctangent, and minus v over (1 plus v squared), which gives minus half a logarithm.
Equations Homogeneous in x and y
So arctan v minus half the logarithm of (1 plus v squared) equals log x plus c, and putting v = y over x and combining the two logarithms gives the printed answer.
One warning. This is the one place in the module where the answer is genuinely not algebraic. A student who expects a polynomial relation and keeps rearranging until one appears will produce something that does not satisfy the equation. The arctangent is the answer.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.