16.1.2 Variables-Separable Equations
| 16.1.2 |
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| 1. Variables-Separable 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”.
Example 1.1
Solve the DE y‘ = xy2.
Solution
y‘ = xy2,
EOS
We can check the correctness of the general solution y = –2/(x2 + C) as follows:
Indeed the general solution is correct.
Separation Of Variables
The DE y‘ = xy2 is called a first-order 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 re-write the given DE as (dy)/y2 = 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 variables-separable, or just separable. Then we integrate both sides of the
resulting equation (dy)/y2 = 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 C1 is an arbitrary constant so is 2C1. Thus for simplicity we can assign 2C1 to C, which consequently is also an
arbitrary constant.
Remark 1.1
Also there must be no differential dx or dy in any denominator.
Relationships Between Solutions
We saw in Section 16.1.1 Part 2 that any 2 solutions of y‘ = f(x) differ by a constant. Adding a non-0 constant to a solution
yields another solution. The equation y‘ = xy2 isn’t of the form y‘ = f(x). Now let’s find out if any 2 different solutions of
y‘ = xy2 differ by a constant, ie if adding a non-0 constant to a solution yields another solution. Let y1 be a solution and K
a non-0 constant. Check to see if y1 + K is also a solution, ie if ( y1 + K)’ = x( y1 + K)2. We have:
So y1 + K isn’t a solution of y‘ = xy2. Adding any non-0 constant to a solution doesn’t yield another solution. No 2
solutions differ by a constant. A solution of y‘ = xy2 is a function whose derivative equals x times the square of the
function itself. Adding a non-0 constant K to a solution y1 generates the function y1 + K whose derivative equals the
derivative of y1 and thus equals x times the square of y1, not x times the square of y1 + K itself, and as a consequence y1
+ K isn’t a solution.
Note that there’s already an arbitrary constant C in the general solution y = –2/(x2 + C). Although no 2 different solutions differ by a non-0 constant and hence adding a non-0 constant to a solution doesn’t yield another solution, the denominators of any 2 different solutions differ by a non-0 constant and hence adding a non-0 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 non-0 constant to a solution yields another solution? Let
y1 be a solution and C a non-0 constant. Then ( y1)’ = f(x) and ( y1 + C)’ = ( y1)’ + 0 = f(x). It follows that y1 + 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
right-hand side f(x) of y‘ = f(x) is a fixed function (not changing into a different function) while the right-hand side xy2 of
y‘ = xy2 is a function that changes into a different function as the solution y changes into a different function.
General Case
A variables-separable (or separable) differential equation is one of the form:
The right-hand 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:
xy2 – x2y2 = (x – x2)y2.
Then we can utilize the variables-separable method to solve the given DE. Note that (x – x2)y2 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 |
Example 2.1
Solve this initial-value problem:
Solution
EOS
In Example 1.1 the general solution y = –2/(x2 + 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 x2 – y2 = 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 x2 – y2 = 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 = x2, y = –(1/2)x2, y = 5x2, etc, ie, all the parabolas y = Cx2 obtained by running the constant C thru
the real numbers, form a family of parabolas. A Family of curves F1 are orthogonal (or perpendicular) to a family of
curves F2 if each curve of F1 is orthogonal (perpendicular) to every curve of F2. 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 = Cx2.
Solution
Differentiating y = Cx2 we get:
x2 + 2y2 = K,
where K = 2K1. Consequently the family of curves orthogonal to the given family of parabolas are ellipses x2 + 2y 2 = K
centered at the origin. See Fig. 3.1.
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EOS
Slope At Any Point (x, y) Of A Parabola y = Cx2. That slope is the derivative dy/dx of y = Cx2 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 = Cx2, 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.
It’s A Problem In Differential Equations
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 variables-separable 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.
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1. Solve the following DEs:
a. dy/dx = x2y 3.
b. du/dr = (sin r)/(cos u).
Solution
sin u = – cos r + C,
sin u + cos r = C.
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2. Solve the following initial-value problems:
Solution
x2 + 2 cot y = 4.
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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 = Ke2x,
where K is an arbitrary constant.
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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.
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5. Show that the solution of the differential equation f ‘(t) = kf(t), where k is a given constant, is f(t) = Aek t, where A is
an arbitrary constant.
Solution
Let y = f(t). Then:
f(t) = Aekt,
where A is an arbitrary constant.
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