Solvable for y
Chapter One Hundred Eight
Syllabus topic Module 2, "2.2 Differential equation of the first order of a degree higher than the first"
Pages 241 to 242 of 303
In one line
Make y the subject, differentiate the whole thing with respect to x, and solve the equation in p and x that results.
MU's label
"Solve for y". The method is longer than factorising and is what to use when factorising fails.
The method
- Rearrange the equation so that y is alone on the left: y = f(x, p).
- Differentiate both sides with respect to x. On the left you get p. On the right you get an expression containing x, p and dp/dx.
- That is a first-order equation in p as a function of x. Solve it by any method of section 2.1.
- You now have a relation between p and x containing the arbitrary constant.
- Eliminate p between that relation and the original equation.
Step five is the one that is forgotten, and without it the answer still contains p, which is not an answer at all.
Why it works
Differentiating y = f(x, p) with respect to x gives, by the chain rule:
p = df/dx + (df/dp)(dp/dx)
The left side is p because dy/dx is p by definition. So the new equation involves only x, p and dp/dx, and y has gone. That is the gain: an equation in two quantities instead of three.
Worked
Solve y = p squared x.
Step one. Already in the right form.
Step two. Differentiate with respect to x. The right side, by the product rule, is 2p x dp/dx plus p squared. Setting that equal to p:
p equals 2p x dp/dx plus p squared.
Step three. Collect: p(1 minus p) equals 2p x dp/dx. Cancelling the p, which assumes p is not zero, gives 1 minus p equal to 2x dp/dx.
That is separable in p and x: dp over (1 minus p) equals dx over 2x.
Integrating: minus log(1 minus p) equals half log x plus c, so 1 minus p equals C over the square root of x, and p equals 1 minus C over root x.
Step four. That is the relation between p and x, with the constant in it.
Step five. Eliminate p by substituting into the original equation. y equals p squared x, so:
y equals (1 minus C over root x) squared times x, which expands to x minus 2C root x plus C squared.
y = p^2 x
y = x - 2 C sqrt(x) + C^2
The checker has substituted that back into the original equation, so the elimination was done correctly.
And the case we cancelled.: p = 0 gives y = 0, which does satisfy the original equation and is not in the general solution at any value of C. So y = 0 is a singular solution, and it was found by noticing what the cancellation threw away.
Solvable for y
A cleaner worked example
Solve y = p x plus p squared, which is Clairaut's form and is covered properly two chapters from here. It is used here because its solvable-for-y working is short and shows every step.
Differentiate: p equals p plus x dp/dx plus 2p dp/dx.
So 0 equals (x plus 2p) dp/dx.
Two cases. Either dp/dx = 0, giving p equal to a constant C; or x plus 2p = 0, giving p equal to minus x over 2.
The first case gives the general solution: substituting p = C back into the original gives y equal to Cx plus C squared.
dy/dx = (-x + sqrt(x^2 + 4y))/2
y = C x + C^2
The second case gives the singular solution: substituting p equal to minus x over 2 gives y equal to minus x squared over 2 plus x squared over 4, which is minus x squared over 4.
dy/dx = -x/2
y = -x^2/4
That split into two cases, one giving the general solution and one the singular, is what always happens with this method, and it is the reason the singular solution exists at all.
The three places it goes wrong
Forgetting that the left side is p. Differentiating y with respect to x gives dy/dx, which is p, not 1 and not dp/dx.
Not eliminating p. An answer containing p is not an answer.
Discarding the second case. The factor that is not dp/dx gives the singular solution, and a question asking for both wants it.
Check yourself
y = p^2 x
y = x - 2 C sqrt(x) + C^2
dy/dx = (-x + sqrt(x^2 + 4y))/2
y = C x + C^2
dy/dx = -x/2
y = -x^2/4
Those three are the two worked examples and the singular solution of the second. Work the first one again from the beginning without looking, and check every step against the five of the procedure; the step people miss is the fifth.
One more point about what this method produces. The relation between p and x at step four sometimes cannot be solved for p, and then p cannot be eliminated in closed form. The honest answer is then the pair of relations, the original equation and the one found at step four, written together. That is a complete answer, it is what a textbook would print, and it is not worth five minutes of an hour trying to force into an explicit form that does not exist.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.