Question
Introduction
This
question was taken from a Facebook group.
The suggested substitution was added in to make it accessible to students
taking H2 Mathematics. In the original
question, no suggested substitution was given.
Strategy
Observe
that the given equation has x
and y. After substitution, we should get an equation
with v and x only.
How to make y “disappear”? One way is recognise patches that can be
substituted for v. Another way is to
make y the subject and differentiate that with
respect to x. Replace the derivative dy/dx with your new
expression. Here, I do both at once!
Solution
Remarks
Usually we try to express y in terms of
x, but here, it is more convenient to express x in terms of
y.
Since A is an arbitrary constant, A+1 is still an arbitrary constant. Also, whenever there is a “ln” appearing in
the solution, it is good to introduce “ln” with an arbitrary constant. So we can lump
A and
1 together and call it ln B. Nothing is lost in this process. ln B is able to achieve all possible numbers. ln B is negative if B is a fraction between 0
and 1, and is positive if B is more than
1 and is zero if B =
1.
This is not
an applied differential equation, so there are no exogenic reasons or hints for
us to suppose that x + y + 2 is a positive quantity. As such, we still need to leave it as | x + y
+ 2|. To get rid of the absolute or
modulus sign properly, we can replace ±B with C.
H04. Look for
pattern(s)
H05. Work
backwards
H09. Restate
the problem in another way
H10. Simplify
the problem
H11. Solve part
of the problem
H13. Use
Equation / write a Mathematical Sentence
Heuristic that cannot be used here
H08. Make
suppositions
Suitable Levels
* GCE ‘A’ Level H2 Mathematics
* IB HL Mathematics Calculus Option
* Advanced Placement (AP) Calculus BC
* university
/ college calculus
* other
syllabuses that involve differential equations
* anyone
who is game for a challenge
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