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**Unformatted text preview: **solution for y . To solve for y we square both sides of (2.4). This gives
2 y + 1 = (ln x + C ) 2 or
y= (ln x + C ) 2 − 1
.
2 Note that constants such as −1 and 2 that appear in the final formula are not arbitrary. They are
specific pieces of the general solution and must not be omitted or changed.
To find the value of C it is more convenient to substitute the initial conditions into (2.4) rather
than into the final expression for y . When we set x = 1 and y = 4 in (2.4) we get C = 3
(assuming the square root is the positive one). The solution of the initial value problem is
therefore y = ((ln x + 3) 2 − 1) / 2 .!
2.3 Money
What is the derivative of a function y = f ( x) ? In Chapter 1 we discussed how to compute
derivatives and in this chapter we learned some methods for finding a function from information
about its derivative. However, to understand how differential equations arise we need to review
the fundamental idea of a derivative as an instantaneous rate of change. You may recall the
standard definition of the derivative in calculus: 22 2 Differential Equations
f ′( x) = lim
h→0 f ( x + h) − f ( x)
.
h (2.5) This definition is useful for doing algebraic exercises and proving the various differentiation rules,
but there is an older, less fashionable notation that captures more of the flavor of the concept.
Instead of using h for the increment in the independent variable x , we use ∆x (read “delta x”).
The symbol ∆x represents a single number, not a product. You can think of ∆x as an
abbreviation for the “difference in x ”. The numerator in (2.5) can then be written as
f ( x + ∆x) − f ( x) = ∆y and represents the difference ∆y in the y value when x changes from x
to x + ∆x . Using the differential notation for the derivative we can rewrite (2.5) as
dy
∆y
= lim
.
dx ∆x →0 ∆x (2.6) The fraction on the right side of (2.6) represents the average rate of change of the function
y = f ( x) over the interval from x to x + ∆x . The derivative is...

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