Worked Separable Equations
Chapter Eighty-Six
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 189 to 190 of 303
In one line
Eight separable equations, worked, every answer substituted back.
One: growth
Solve dy/dx = 5y.
Separating gives dy over y equal to 5 dx, so log of the size of y is 5x plus c.
dy/dx = 5y
y = C e^(5x)
Two: decay towards a value
Solve dy/dx = 4 minus y.
Separating gives dy over (4 minus y) equal to dx, so minus log of the size of (4 minus y) is x plus c, and hence 4 minus y is C e to the minus x.
dy/dx = 4 - y
y = 4 + C e^(-x)
The minus sign from integrating one over (4 minus y) is the step that is dropped most often. Check it by differentiating your answer, which takes five seconds.
Three: a product on the right
Solve dy/dx = x squared y cubed.
Separating gives dy over y cubed equal to x squared dx, so minus one over 2y squared equals x cubed over 3 plus c. Multiplying through by minus 6 and gathering the constant on the right gives the answer below.
dy/dx = x^2 y^3
3/y^2 + 2 x^3 = C
That implicit form is tidier than the explicit one, which would carry a square root and a sign. Multiplying out and renaming the constant is always allowed and is worth doing when it removes a fraction. Note that the constant is gathered alone on one side: an implicit answer is a statement that some function of x and y is constant, and writing it any other way makes it harder to check.
Four: a sum that factorises
Solve dy/dx = xy plus x.
The right-hand side factorises as x(y plus 1), so it separates after all.
dy/dx = x y + x
y = -1 + C e^(x^2/2)
Try factorising before declaring an equation non-separable. This one looks like a sum and is a product.
Five: an exponential of a sum
Solve dy/dx = e to the (2x minus y).
The right side is e to the 2x times e to the minus y.
dy/dx = e^(2x - y)
e^y = e^(2x)/2 + C
Six: a trigonometric quotient
Solve dy/dx = cos x over y.
Separating gives y dy equal to cos x dx.
dy/dx = cos(x)/y
y^2 = 2 sin(x) + C
Seven: an answer that will not be made explicit
Solve dy/dx = (1 plus y squared) over y.
Separating gives y dy over (1 plus y squared) equal to dx, so half the logarithm of (1 plus y squared) is x plus c. Exponentiating and gathering the constant gives the answer below.
dy/dx = (1 + y^2)/y
(y^2 + 1) e^(-2x) = C
Worked Separable Equations
Equivalently y squared equals C e to the 2x minus 1, which is the form most students would write. Both say the same thing; the first has the constant alone on one side, which is what makes it checkable in one step.
Eight: with an initial condition
Solve dy/dx = y squared, with y(0) = 1.
Separating gives dy over y squared equal to dx, so minus one over y equals x plus c, and y equals minus one over (x plus c).
dy/dx = y^2
y = -1/(x + C)
At x = 0 with y = 1: 1 equals minus one over c, so c is minus 1.
dy/dx = y^2
y = 1/(1 - x)
That particular solution is worth a moment. It grows without limit as x approaches 1, and beyond x = 1 it is a different branch. So the solution exists only on an interval, not for all x, even though the equation looked perfectly harmless. A non-linear equation can have a solution that blows up in finite time, and a linear one never can. That is one of the real differences between the two.
The pattern across the eight
| Right-hand side | Integrating gives | Answer contains |
|---|---|---|
| a multiple of y | a logarithm | an exponential |
| a constant minus y | a logarithm | a constant plus an exponential |
| a power of y | a power of y | an implicit relation |
| a factorisable sum | a logarithm | an exponential, shifted |
The first two rows are the ones that arise from the models of the earlier chapter, and they are the two shapes to recognise instantly.
Check yourself
dy/dx = -y/2
y = C e^(-x/2)
dy/dx = x/(y + 1)
(y + 1)^2 = x^2 + C
dy/dx = y - y^2
y = 1/(1 + C e^(-x))
The third is the logistic equation of the modelling chapter, with its maximum set to 1. Separating it needs partial fractions on one over y(1 minus y), which is the first place in this module where a technique from Module 1 is reused.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.