Bernoulli's Equation, and Other Equations Reducible to the Linear Form
Chapter One Hundred Two
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 226 to 228 of 303
In one line
Divide by y to the n and substitute for y to the power one minus n, and Bernoulli's equation becomes linear.
The shape
dy/dx + P y = Q y^n
It is the linear equation with a power of y on the right instead of a function of x alone. For n = 0 it is linear; for n = 1 it is separable; for every other n it is neither, and this method is what it needs.
MU's label is "Linear Equation and equation reducible to this form", and this is the principal thing that reduces.
The method
- Divide the whole equation by y to the n.
- Put v equal to y to the power (1 minus n).
- Then dv/dx is (1 minus n) times y to the minus n times dy/dx, which is exactly what the divided equation contains.
- The equation becomes linear in v, and the previous chapter solves it.
- Substitute back.
Why that substitution
Dividing by y to the n gives y to the minus n dy/dx, plus P y to the power (1 minus n), equal to Q.
The second term is P times y to the (1 minus n), which suggests calling that v. And then dv/dx is (1 minus n) y to the minus n dy/dx, so the first term is dv/dx over (1 minus n). Both terms are now in v, and the equation is linear.
dv/dx/(1 - n) + P v = Q
Multiplying by (1 minus n) puts it in standard form with P replaced by (1 minus n)P and Q by (1 minus n)Q.
Worked
Solve dy/dx plus y = y squared.
Here P is 1, Q is 1 and n is 2, so v is y to the power minus 1, that is one over y.
Dividing by y squared: y to the minus 2 dy/dx plus one over y equals 1.
With v equal to one over y, dv/dx is minus y to the minus 2 dy/dx, so the first term is minus dv/dx.
The equation becomes minus dv/dx plus v equals 1, that is dv/dx minus v equals minus 1.
That is linear with P equal to minus 1, so mu is e to the minus x, and the solution is v e to the minus x equal to the integral of minus e to the minus x, which is e to the minus x plus C.
So v equals 1 plus C e to the x, and y is one over that.
dy/dx + y = y^2
y = 1/(1 + C e^x)
That is the logistic equation of the modelling chapter, with the signs arranged differently, and the answer is the same S-shaped family.
Bernoulli's Equation, and Other Equations Reducible to the Linear Form
Worked, with a power of x
Solve dy/dx plus y over x = y squared.
P is one over x, Q is 1, n is 2, so v is one over y.
Dividing and substituting as before gives minus dv/dx plus v over x equals 1, that is dv/dx minus v over x equals minus 1.
Linear with P equal to minus one over x, so mu is one over x. The solution is v over x equal to the integral of minus one over x, which is minus log x plus C.
So v equals x(C minus log x), and y is its reciprocal.
dy/dx + y/x = y^2
y = 1/(x(C - log(x)))
Worked, with n = 3
Solve dy/dx plus y = y cubed.
n is 3, so v is y to the power minus 2. Dividing by y cubed and substituting, with dv/dx equal to minus 2 y to the minus 3 dy/dx, gives minus dv/dx over 2 plus v equals 1, that is dv/dx minus 2v equals minus 2.
Linear with P equal to minus 2, so mu is e to the minus 2x, and the solution is v e to the minus 2x equal to the integral of minus 2 e to the minus 2x, which is e to the minus 2x plus C.
So v equals 1 plus C e to the 2x, and y squared is one over that.
dy/dx + y = y^3
y = 1/sqrt(1 + C e^(2x))
The two other things that reduce to linear form
Linear in x rather than y. Turn the equation upside down, as the previous chapter's last worked example did. This is worth trying on anything that is not linear in y.
An equation in f(y). If the equation contains only a particular function of y and its derivative, substitute for that function. Putting v equal to e to the y turns an equation in e to the y and its derivative into a linear one, because dv/dx is e to the y dy/dx.
dy/dx = e^(-y)
e^y = x + C
With v equal to e to the y that equation reads dv/dx equal to 1, so v is x plus C, which is the answer. It is separable as well, and either route is fine.
Recognising which method
| Equation | n | Method |
|---|---|---|
| dy/dx + Py = Q | 0 | linear |
| dy/dx + Py = Qy | 1 | separable, after collecting |
| dy/dx + Py = Qy^2 | 2 | Bernoulli, v = 1/y |
| dy/dx + Py = Qy^3 | 3 | Bernoulli, v = 1/y^2 |
| dy/dx + Py = Q sqrt(y) | 1/2 | Bernoulli, v = sqrt(y) |
The last row is worth noting: n need not be a whole number, and a square root of y on the right is a Bernoulli equation with n = one half, so v is y to the power one half.
Bernoulli's Equation, and Other Equations Reducible to the Linear Form
Check yourself
dy/dx - y = y^2
y = -1/(1 + C e^(-x))
dy/dx + 2y/x = y^2
y = 1/(x^2(C + 1/x))
For the second, v is one over y and the linear equation in v is dv/dx minus 2v over x equal to minus 1, whose factor is x to the minus 2.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.