Separating the Variables
Chapter Eighty-Five
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 187 to 188 of 303
In one line
If the equation can be written with all the x on one side and all the y on the other, integrate both sides.
MU's first label
"Separation of variables". It is the first method of section 2.1 and the one every other method tries to reduce to.
When it applies
The equation must be writable in the form below, with the two variables completely separated.
f(y) dy = g(x) dx
Equivalently, dy/dx must be a product or a quotient of a function of x alone and a function of y alone.
| Equation | Separable | Why |
|---|---|---|
| dy/dx = x y | yes | a product |
| dy/dx = x/y | yes | a quotient |
| dy/dx = x + y | no | a sum cannot be split |
| dy/dx = e to the (x + y) | yes | the exponential of a sum is a product |
| dy/dx = (x y plus x)/(y) | yes | factorise the top as x(y plus 1) |
The fourth row is the trick worth knowing: a sum in an exponent is a product of exponentials, so such an equation is separable even though it does not look it.
The fifth row is the other one: factorise before deciding. Many equations that look like sums separate after one factorisation.
The method
- Write the equation as f(y)dy = g(x)dx.
- Integrate both sides.
- Put one arbitrary constant, on one side only.
- Tidy, and make the answer explicit if it easily can be.
- Note anything you divided by, and check whether it gives a lost solution.
Step three is worth a sentence. Both integrations produce a constant, but the difference of two arbitrary constants is one arbitrary constant, so only one is written. Writing two is not wrong, only untidy.
Worked: the simplest
Solve dy/dx = 2xy.
Separate: dy over y equals 2x dx.
Integrating: log of the size of y equals x squared plus c.
Exponentiating: the size of y equals e to the c times e to the x squared, and writing C for plus or minus e to the c gives the answer.
dy/dx = 2xy
y = C e^(x^2)
Note the step where the constant became a multiplier. Integrating gave a constant added to a logarithm, and exponentiating turned it into a factor. That happens in nearly every separable equation with a logarithm, and writing C for e to the c is standard.
Worked: a quotient
Solve dy/dx = x over y.
Separate: y dy equals x dx. Integrating: y squared over 2 equals x squared over 2 plus c, so y squared minus x squared equals C.
dy/dx = x/y
y^2 - x^2 = C
The answer is implicit, and it is a family of hyperbolas. Making it explicit would mean a square root and a sign choice, and the implicit form is better.
Separating the Variables
Worked: with the exponential trick
Solve dy/dx = e to the (x plus y).
The right-hand side is e to the x times e to the y, so it is separable after all.
Separate: e to the minus y dy equals e to the x dx. Integrating: minus e to the minus y equals e to the x plus c.
dy/dx = e^(x + y)
e^(-y) + e^x = C
Worked: a trigonometric one
Solve dy/dx = y cos x.
Separate: dy over y equals cos x dx. Integrating: log of the size of y equals sin x plus c.
dy/dx = y cos(x)
y = C e^(sin(x))
Worked: with an initial condition
Solve dy/dx = minus 2xy with y(0) = 3.
The general solution, by the first worked method with a sign change, is C e to the minus x squared.
dy/dx = -2xy
y = C e^(-x^2)
At x = 0 the exponential is 1, so C = 3 and the particular solution is 3 e to the minus x squared.
dy/dx = -2xy
y = 3 e^(-x^2)
The solution you may have divided away
In the very first example, separating meant dividing by y, which is illegal when y = 0. And y = 0 is a solution: both sides of dy/dx = 2xy are then zero.
Here it is recovered by taking C = 0, so nothing was lost. But the check should be made, not assumed. The habit: write down what you divided by, and test whether it being zero solves the equation.
The commonest mistakes
A constant on both sides. One is enough, and two make the tidying harder.
Forgetting the modulus in a logarithm. The integral of one over y is the logarithm of the size of y. Dropping it is usually harmless because the constant absorbs the sign, but it should be written.
Separating something that does not separate. dy/dx = x plus y is not separable and no amount of rearranging makes it so. It is linear, and the chapter on the linear equation solves it.
Check yourself
dy/dx = 3y
y = C e^(3x)
dy/dx = y/x
y = C x
dy/dx = x y^2
y = -2/(x^2 + C)
dy/dx = (1 + y^2)/(1 + x^2)
y = (x + C)/(1 - C x)
The third is non-linear and separable, and the answer has the constant inside the bracket rather than as a multiplier, which is what happens when the integration does not produce a logarithm. The fourth comes out as arctan y equals arctan x plus c, and taking the tangent of both sides with the compound-angle formula gives the printed form.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.