Worked Equations Solvable for p
Chapter One Hundred Seven
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first"
Pages 239 to 240 of 303
In one line
Six worked examples, including one that needs the quadratic formula and one whose factors are equations of two different kinds.
One: both roots constant
Solve p squared minus 4 = 0.
Factorising: (p minus 2)(p plus 2) = 0.
p^2 - 4 = 0
y = 2x + C
p^2 - 4 = 0
y = -2x + C
General solution: the product of (y minus 2x minus C) and (y plus 2x minus C), set to zero.
Two: one root zero
Solve p squared minus 3p = 0.
Factorising: p(p minus 3) = 0, so p is 0 or 3.
p^2 - 3p = 0
y = C
p^2 - 3p = 0
y = 3x + C
The root p = 0 gives y equal to a constant, the family of horizontal lines. That is a perfectly good solution and students sometimes discard it as trivial, losing the mark.
Three: roots involving x
Solve p squared minus (x plus 1)p plus x = 0.
The roots are x and 1, so it factorises as (p minus x)(p minus 1) = 0.
p - x = 0
y = x^2/2 + C
p - 1 = 0
y = x + C
Four: factors of two different kinds
Solve x p squared minus (x squared plus 1)p plus x = 0.
Dividing by x and looking for roots whose product is 1 and whose sum is x plus one over x: those are x and one over x.
So it factorises as (p minus x)(p minus one over x) = 0.
p - x = 0
y = x^2/2 + C
p - 1/x = 0
y = log(x) + C
One factor integrated directly, the other gave a logarithm. Different kinds of answer from one equation is normal.
Five: needing the formula
Solve p squared plus 2p y cot x = y squared.
Treat it as a quadratic in p, with a equal to 1, b equal to 2y cot x and c equal to minus y squared.
The discriminant is 4y squared cot squared x plus 4y squared, which is 4y squared(cot squared x plus 1), which is 4y squared cosec squared x, a perfect square.
So p equals minus y cot x plus or minus y cosec x, that is y(minus cos x plus or minus 1) over sin x.
First root, with the plus: p equals y(1 minus cos x) over sin x. Using the half-angle identities, (1 minus cos x) over sin x is tan(x over 2), so p equals y tan(x over 2).
Separating: dy over y equals tan(x over 2) dx, so log y equals minus 2 log(cos(x over 2)) plus c.
Worked Equations Solvable for p
p - y tan(x/2) = 0
y = C/cos(x/2)^2
Second root, with the minus: p equals minus y(1 plus cos x) over sin x, which is minus y cot(x over 2).
p + y/tan(x/2) = 0
y = C/sin(x/2)^2
Two families, and notice how similar they are: one is a constant over cos squared of x over 2, the other a constant over sin squared. Their ratio is tan squared of x over 2, which is the tidy way to see that they are genuinely different families.
The lesson from this one: when the discriminant turns out to be a perfect square, the trigonometric identities are usually what make it so. Look for cot squared plus 1 equal to cosec squared, and for the half-angle forms.
Six: a cubic
Solve p cubed minus 2 p squared minus p plus 2 = 0.
Trying p = 1 gives 1 minus 2 minus 1 plus 2 = 0, so (p minus 1) is a factor. Dividing out gives p squared minus p minus 2, which factorises as (p minus 2)(p plus 1).
So the three roots are 1, 2 and minus 1.
p^3 - 2 p^2 - p + 2 = 0
y = x + C
p^3 - 2 p^2 - p + 2 = 0
y = 2x + C
p^3 - 2 p^2 - p + 2 = 0
y = -x + C
Three families, one per root, exactly as the degree promised.
What to check before moving on
One family per root. A quadratic gives two, a cubic three.
Each family satisfies the original equation. Substitute and see.
The constant is the same letter throughout. One arbitrary constant, however many families.
Check yourself
p^2 - 16 = 0
y = 4x + C
p^2 - 5p = 0
y = 5x + C
p^2 - x p = 0
y = x^2/2 + C
p^2 - (2x + 1) p + 2x = 0
y = x^2 + C
For the last, the roots are 2x and 1, so the other family is y equal to x plus C.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.