Worked Homogeneous Equations
Chapter Eighty-Nine
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 196 to 198 of 303
In one line
Five homogeneous equations worked in full, including the partial-fraction integration the v equation usually needs.
One: the simplest
Solve dy/dx = (y minus x) over x.
Both parts are of degree 1. Put y = vx, so dy/dx is v plus x dv/dx, and the right-hand side becomes v minus 1.
So x dv/dx equals minus 1, giving dv equal to minus dx over x, and v equals minus log x plus c.
dy/dx = (y - x)/x
y = -x log(x) + C x
Two: where the v equation needs partial fractions
Solve dy/dx = (y squared) over (xy minus x squared).
Degree 2 on top and on the bottom, so homogeneous. Dividing top and bottom by x squared gives v squared over (v minus 1).
So v plus x dv/dx equals v squared over (v minus 1), and x dv/dx equals v squared over (v minus 1) minus v, which is v over (v minus 1).
Separating: (v minus 1) dv over v equals dx over x. The left side is 1 minus 1 over v, which integrates to v minus log v.
So v minus log v equals log x plus c, and putting v = y over x gives y over x minus log(y over x) equals log x plus c, that is y over x minus log y equals c.
dy/dx = y^2/(x y - x^2)
y/x - log(y) = C
The logarithms combined at the last step, which is why the x disappeared. Watching for that combination is worth a minute: it usually tidies the answer considerably.
Three: a quotient with a square root
Solve dy/dx = (y plus sqrt(x squared plus y squared)) over x, for positive x.
Every term is of degree 1, including the square root, because the square root of a degree-2 expression is of degree 1. So it is homogeneous.
Putting y = vx, the right-hand side becomes v plus the square root of (1 plus v squared), so the v cancels on subtraction and x dv/dx equals sqrt(1 plus v squared).
Separating: dv over sqrt(1 plus v squared) equals dx over x. The left side is the standard integral that gives an inverse hyperbolic sine, which Module 1's chapter on the inverse hyperbolic functions derived.
So sinh inverse of v equals log x plus c, and putting v = y over x gives the answer.
dy/dx = (y + sqrt(x^2 + y^2))/x
asinh(y/x) - log(x) = C
Equivalently, taking the sinh of both sides, y over x equals sinh(log x plus c), which can be expanded into an algebraic form; the implicit answer above is the one to write.
This is the first place in Module 2 where a result from Module 1 is used directly, and it is worth noticing: the two halves of the paper are not separate subjects.
Worked Homogeneous Equations
Four: with the x and y roles reversed
Solve dy/dx = (x squared plus y squared) over (x squared).
Homogeneous of degree 2 over degree 2. Putting y = vx gives v plus x dv/dx equal to 1 plus v squared, so x dv/dx equals 1 minus v plus v squared.
The quadratic 1 minus v plus v squared has negative discriminant, so completing the square gives an arctangent.
dy/dx = (x^2 + y^2)/x^2
2 atan((2y/x - 1)/sqrt(3))/sqrt(3) - log(x) = C
The root three comes from the completed square, and it is the signature of an irreducible quadratic in v. Expect it rather than suspecting a mistake.
Five: an equation that looks homogeneous and is not
Solve dy/dx = (x squared plus y) over x.
The top has a degree-2 term and a degree-1 term, so it is not homogeneous. Try the substitution anyway and see what goes wrong: putting y = vx gives v plus x dv/dx equal to (x squared plus vx) over x, which is x plus v, so x dv/dx equals x. The v has cancelled but so has everything else, and dv equals dx, giving v equal to x plus c.
That happens to work here, and the answer is y equal to x squared plus cx. But it worked by luck: the substitution left an equation in x alone, which is not what the method promises, and on a genuinely non-homogeneous equation it leaves a mixture of x and v that separates for neither.
dy/dx = (x^2 + y)/x
y = x^2 + C x
The honest route is the linear one: the equation is dy/dx minus y over x equal to x, which is linear of the first order, and the chapter on the linear equation solves it in three lines with no guessing.
The test to apply before substituting is the one at the top of the previous chapter: divide out and see whether the right-hand side is a function of y over x alone. Here it is x plus y over x, which contains a bare x, so it is not.
The routine, condensed
- Check the degrees.
- Put y = vx and dy/dx = v plus x dv/dx.
- Simplify to x dv/dx equal to something in v alone. Every x must cancel.
- Separate: that something's reciprocal times dv equals dx over x.
- Integrate. Expect a logarithm on the right, and on the left either a logarithm, an arctangent, or a partial fraction.
- Put v = y over x back.
- Combine logarithms if you can.
Worked Homogeneous Equations
Check yourself
dy/dx = (2y - x)/x
y = x + C x^2
dy/dx = (x + 2y)/x
y = -x + C x^2
dy/dx = y/x + y^2/x^2
x/y + log(x) = C
The third needed partial fractions on the v side, and the answer combines into a tidy relation between x over y and log x.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.