Worked Equations Solvable for y
Chapter One Hundred Nine
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first"
Pages 243 to 244 of 303
In one line
Four worked examples, and the observation that most questions of this kind turn out to be Clairaut's form in disguise.
One: y = p squared x
Done in the previous chapter and repeated because it is the cleanest illustration.
Differentiating gives p equal to 2px dp/dx plus p squared. Cancelling p and separating gives 1 minus p equal to 2x dp/dx, so p equals 1 minus C over root x, and substituting back gives the answer.
y = p^2 x
y = x - 2 C sqrt(x) + C^2
And the cancelled case p = 0 gives the singular solution.
dy/dx = 0
y = 0
Two: y = 4 p squared x
The same work with a constant in it, which is worth doing once so that the constant's effect is visible.
Differentiating: p equals 8px dp/dx plus 4p squared. Cancelling p: 1 minus 4p equals 8x dp/dx.
Separating and integrating gives 1 minus 4p equal to C over root x, so p equals (1 minus C over root x) over 4.
Substituting into y = 4p squared x:
y = 4 p^2 x
y = (x - 2 C sqrt(x) + C^2)/4
The whole answer is the first example's divided by 4, which is what the constant did.
Three: y = x p plus one over p
Make y the subject: it already is.
Differentiating: p equals p plus x dp/dx minus (1 over p squared) dp/dx.
So 0 equals (x minus one over p squared) dp/dx, which splits into two cases.
Case one, dp/dx = 0, so p is a constant C, and substituting gives the general solution.
y = x p + 1/p
y = C x + 1/C
Case two, x equals one over p squared, so p equals one over root x, and substituting gives y equal to root x plus root x, which is 2 root x.
y = x p + 1/p
y = 2 sqrt(x)
That second answer is a singular solution: no value of C in y = Cx plus one over C gives y = 2 root x, because the general solution is a family of straight lines and this is a parabola. Squaring it gives y squared = 4x.
Four: the observation worth having
Look at the third example. Its form was y = xp plus a function of p alone, which is Clairaut's form, and the method collapsed into two cases at once: dp/dx = 0 giving the general solution, and the other factor giving the singular one.
That is not a coincidence about that example. Most equations set as solvable-for-y questions are Clairaut's form or reduce to it, because those are the ones whose differentiated equation factorises neatly. The first two examples were not, and notice how much more work they took.
Worked Equations Solvable for y
So the practical advice: before grinding through the solvable-for-y method, check whether the equation is y = xp plus f(p). If it is, the Clairaut chapter gives the answer in one line.
When the answer is parametric
Sometimes step four gives a relation between p and x that cannot be solved for p, and then p cannot be eliminated.
Take y = 2px plus p squared. Differentiating gives p equal to 2p plus 2x dp/dx plus 2p dp/dx, so minus p equals 2(x plus p) dp/dx.
Turning that upside down, dx/dp equals minus 2(x plus p) over p, which rearranges to a linear equation in x with p as the variable: dx/dp plus 2x over p equals minus 2.
Its integrating factor is p squared, and solving gives x equal to minus 2p over 3 plus C over p squared.
Substituting into the original gives y equal to minus p squared over 3 plus 2C over p.
So the answer is that pair of equations, x and y each in terms of p and C. The parameter p cannot be eliminated in closed form, and the pair is the answer. That is a parametric solution, it is complete, and a question whose working leads there is testing whether you know to stop.
The procedure, once more
- Make y the subject.
- Differentiate with respect to x, remembering that the left side is p.
- Solve the resulting equation in p and x. If it is easier as an equation for x in terms of p, turn it upside down, which often makes it linear.
- Eliminate p between that and the original equation.
- If p cannot be eliminated, give the parametric pair.
- Check the case you cancelled: it usually gives the singular solution.
Check yourself
y = p^2 x
y = x - 2 C sqrt(x) + C^2
y = x p + 1/p
y = C x + 1/C
y = x p + 1/p
y = 2 sqrt(x)
The second and third are the general and the singular solution of the same equation, and a question asking to "solve completely" wants both.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.