SOLUTIONS:
(a) By differentiating equation x^2 + y^2 = r^2 implicitly, we get 2x + 2y from which
which is the given d.e.
(b) x^2 + y^2 = r^2 describes circles centered at the origin. For initial point (4,−3), (4)^2 + (−3)^2 =- So x^2 + y^2 = 25. However, this is not the particular solution.
A particular solution must be differentiable on an interval containing the initial point. This
circle is not differentiable at (−5,0) and (5,0). (The d.e. shows undefined when y = 0, and
the slope field shows vertical tangents along the x-axis.) Hence, the particular solution
includes only the semicircle in quadrants III and IV.
Solving x^2 + y^2 = 25 for y yields The particular solution through point (4,−3) is
with domain −5 < x < 5.
Derivatives of Implicitly Defined Functions
In Examples 2 and 3 above, each d.e. was of the form = f (x) or y ′ = f (x). We were able to find the
general solution in each case very easily by finding the antiderivative
We now consider d.e.’s of the form where f (x,y) is an expression in x and y; that is,
is an implicitly defined function. Example 4 illustrates such a differential equation. Here is another
example.
EXAMPLE 5
Figure N9–5 shows the slope field forAt each point (x,y) the slope is the sum of its coordinates. Three particular solutions have been
added, through the points