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1. VariablesSeparable Differential Equations 
All the sections in Chapter 12 concentrated on
applications of the definite integral. In this chapter we'll focus on an
application
of the indefinite integral, the one to differential equations. In the remainder
of this section the abbreviation “DE” stands for
“differential equation”.
Solve the DE y' = xy^{2}.
y' = xy^{2},
EOS
We can check the correctness of the general solution y = –2/(x^{2} + C) as follows:
Indeed the general solution is correct.
The DE y' = xy^{2} is called a firstorder
differential equation because it involves a derivative of the first order
and none
of higher order.
We write the derivatives in the notation dy/dx if they're not given in that notation. We
regard dx and dy as differentials
and thus dy/dx as a normal fraction. Then
we rewrite the given DE as (dy)/y^{2}
= x dx. By doing so we separate
its
variables by putting all terms involving y
on the left and all terms involving x
on the right. Because such a separation is
possible, the given DE is said to be variablesseparable, or just separable.
Then we integrate both sides of the
resulting equation (dy)/y^{2}
= x dx and get the general
solution of the given DE. This integration is valid because if 2
functions are equal then their general antiderivatives are equal up to a
constant.
As C_{1} is an
arbitrary constant so is 2C_{1}. Thus for
simplicity we can assign 2C_{1} to C, which consequently is also
an
arbitrary constant.
Also there must be no differential dx or dy in any denominator.
We saw in Section
16.1.1 Part 2 that any 2 solutions of y' = f(x) differ by a constant.
Adding a non0 constant to a solution
yields another solution. The equation y' = xy^{2} isn't of the
form y' = f(x). Now let's find out if any
2 different solutions of
y' = xy^{2} differ by a
constant, ie if adding a non0 constant to a solution yields another solution.
Let y_{1}
be a solution and K
a non0 constant. Check to see if y_{1} + K is also a solution, ie if ( y_{1} + K)' = x( y_{1}
+ K)^{2}. We have:
So y_{1}
+ K isn't a
solution of y' = xy^{2}. Adding any
non0 constant to a solution doesn't yield another solution. No 2
solutions differ by a constant. A solution of y' = xy^{2} is a function whose derivative
equals x times the
square of the
function itself. Adding a non0 constant K
to a solution y_{1} generates the
function y_{1} + K whose derivative equals the
derivative of y_{1} and thus equals
x times the square of y_{1}, not x times the square of y_{1} + K
itself, and as a consequence y_{1}
+ K isn't a
solution.
Note that there's already an arbitrary constant C in the general solution y = –2/(x^{2} + C).
Although no 2 different
solutions differ by a non0 constant and hence adding a non0 constant to a
solution doesn't yield another solution, the
denominators of any 2 different solutions differ by a non0 constant and hence
adding a non0 constant to the
denominator of a solution does yield another solution.
Looking back at the DE of the form y' = f(x), why does adding a non0 constant to a
solution yields another solution? Let
y_{1} be a solution and C a non0 constant. Then ( y_{1})' = f(x) and ( y_{1} + C)' = ( y_{1})' + 0 = f(x). It follows that y_{1} + C is a
solution too. Therefore the answer is because the derivative of each is f(x), as required by the DE. Remark that the
righthand side f(x) of y' = f(x) is a fixed function (not changing into a
different function) while the righthand side xy^{2} of
y' = xy^{2} is a function
that changes into a different function as the solution y changes into a different function.
A variablesseparable (or separable) differential equation is one of the form:
The righthand side is in the form F(x,
y) (function of 2 variables x and y) but not in the form f(x)g( y) ( f(x) times
g( y)). However it can be brought into the latter
form by factoring:
xy^{2} – x^{2}y^{2} = (x – x^{2})y^{2}.
Then we can utilize the variablesseparable method to solve
the given DE. Note that (x
– x^{2})y^{2} is in the form f(x)g(
y) and
also in the form F(x, y).
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2. Explicit And Implicit Solutions 
Solve this initialvalue problem:
EOS
In Example 1.1 the general
solution y = –2/(x^{2} + C) expresses the general solution y explicitly as a function of x (ie in
the form y = f(x)), whereas in this example above the general
solution x^{2} – y^{2} = C expresses the general solution y
implicitly as a function of x
(ie not in the form y
= f(x)). Solutions of certain DE's
are often expressed in implicit form for
convenience, or sometimes by necessity due to the difficulty involved in
obtaining an explicit form. Note that the particular
solution x^{2} – y^{2} = 5 also expresses the
particular solution y
implicitly as a function of x.
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3. Orthogonal Families Of Curves 
The parabolas y
= x^{2}, y = –(1/2)x^{2}, y
= 5x^{2}, etc, ie, all
the parabolas y = Cx^{2} obtained by running the
constant C thru
the real numbers, form a family of parabolas. A Family of curves F_{1} are orthogonal
(or perpendicular) to a family of
curves F_{2} if each curve
of F_{1} is orthogonal
(perpendicular) to every curve of F_{2}. See Fig. 3.1
for an example. The
explanation of some points other than the method of variables separable in the
solution of the following example follows
immediately the solution.
Example 3.1
Find the family of curves orthogonal to the family of parabolas y = Cx^{2}.
Solution
Differentiating y = Cx^{2} we get:
x^{2} + 2y^{2} = K,
where K
= 2K_{1}. Consequently
the family of curves orthogonal to the given family of parabolas are ellipses x^{2} + 2y ^{2} = K
centered at the origin. See Fig. 3.1.
Fig. 3.1
Family of curves y = Cx^{2} are orthogonal to family of

EOS
Slope At Any Point (x, y) Of A Parabola y = Cx^{2}. That slope is the derivative dy/dx of y = Cx^{2} at the point (x, y).
Elimination Of The Constant C. The constant C is eliminated in the calculation of the
derivative dy/dx of y
= Cx^{2}, and
dy/dx is expressed in terms of both x and y. This is because an orthogonal curve would be
orthogonal to every parabola
of the given family, not just one particular parabola.
Slope At Any Point (x, y)
Of An Orthogonal Curve. That slope is the negative reciprocal of the slope
at the point (x, y) of
the parabola that passes thru that point.
We see that the problem of finding a family of curves
orthogonal to a given family of curves is a problem in differential
equations. We can sometimes use the variablesseparable method to solve it.
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4. Differential Equations And Integration 
The DEs in Section
16.1.1 are of the form y’ = f (x)
and y’’ = f (x).
To solve them, we integrate f (x) once and twice respectively.
The DEs in this section are of the form dy/dx = f (x, y) or y' = f (x, y).
To solve them, we first perform some algebraic
manipulation, then we integrate.
Problems & Solutions 
1. Solve the following DEs:
a. dy/dx = x^{2}y ^{3}.
b. du/dr = (sin r)/(cos u).
Solution
sin u = – cos r + C,
sin u + cos r = C.
2. Solve the following initialvalue problems:
Solution
x^{2} + 2 cot y = 4.
3. Find the family of curves orthogonal to the family x + y ^{2} = C. Sketch some members of each of the families.
Solution
Differentiating x + y ^{2} = C implicitly with respect to x we get:
y = Ke^{2}^{x},
where K is an arbitrary constant.
4. Suppose an object of mass m falling freely near the
surface of the Earth is retarded by a force of air resistance of
magnitude proportional to its velocity. So the force of air resistance is
kv, where k > 0 is a constant and v = v(t) is
the velocity of the object at time t. Then, according to Newton's
second law of motion:
Solution
c. Yes, because e^{–}^{kt}^{/}^{m} decreases to 0 rapidly as t increases.
d. Velocity becomes constant when a = 0, so when mg – kv = 0 or v = mg/k.
5. Show that the solution of the differential
equation f '(t)
= kf(t), where k is a given constant, is f(t)
= Ae^{k}^{ }^{t},
where A is
an arbitrary constant.
Solution
Let y = f(t). Then:
f(t) = Ae^{kt},
where A is an arbitrary constant.
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