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CALCULUS 1 PROBLEMS & SOLUTIONS

7.8
Differential Equations – Variables Separable


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Review


1. Note

All the previous sections in this chapter concentrated on applications of the definite integral. In this
section we'll focus on an application of the indefinite integral, the one to differential equations. The
subject of differential equations was first introduced in Section 3.9.

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2. Variables–Separable Differential Equations

Consider the equation x² – y ² = 1. Differentiating it implicitly with respect to x we get 2x – 2y(dy/dx) =
0, so that dy/dx = x/y, which is a differential equation.

Now, to work in reverse, suppose we're given the differential equation:



where C = –2C₁. The curve x² – y ² = C is the general solution of the given equation. Clearly, the curve
x² – y ² = 1 is a particular solution.

Separating Variables. In re-writing the given equation as y dy = x dx, 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 equation is said to be variables-separable, or just separable. Then we integrate
both sides of the resulting equation y dy = x dx and obtain the solution of the given equation.


General Case

A variables-separable (or separable) differential equation is one of the form:




Remark 1

ò (1/y ²) dy = –1/y + C because (d/dy)(–1/y + C) = –(–1/y²) + 0 = 1/y², but ò (1/y ²) dx ¹ –1/y + C
because (d/dx)(–1/y + C) = –(–1/y²)(dy/dx) + 0 = (1/y²)(dy/dx) ¹ 1/y². It's important that all terms
containing x are with the differential dx on one side and that all terms containing y are with the
differential dy on the other side. For example, in ò (1/g( y)) dy, y is the variable with respect to which
we integrate; here it's not a function. If you run into ò (1/g( y)) dx or ò f(x) dy while you're intending to
utilize the variables-separable method, back off and try again until you get only ò (1/g( y)) dy and
ò f(x) dx.

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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 each
curve of F₂. See Fig. 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 2

Find the family of curves orthogonal to the family of parabolas y = Cx².

Solution
Differentiating y = Cx² we get:



where K = 2K₁. Thus, the family of curves orthogonal to the given family of parabolas are ellipses
x² + 2y ² = K centered at the origin. See Fig. 1.

Fig. 1 Family of curves y = Cx^2 are orthogonal to family of curves x^2 + 2y^2 = K.

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 each 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 passing 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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Problems & Solutions


1. Solve the following differential equations:

a. dy/dx = x²y ³. b. du/dr = (sin r)/(cos u).

Solution



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2. Solve the following initial-value problems.

a. dy/dx = xy ³, y(0) = 1. b. y' – x sin² y = 0, y = p/2 when x = 2.

Solution



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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:



where K is an arbitrary non-0 constant. For K = 0, we have y = 0, for which we obtain dy/dx = 0 =
2 . 0 = 2y. Thus, y = 0 is also a solution of dy/dx = 2y. Consequently, the orthogonal family is y =
Ke^2x, 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:



where a is the acceleration of the object at time t and g is the acceleration of gravity. Assume the
object falls from rest at time t = 0.

a. Find the velocity v(t) at any time t > 0.
b. Show that velocity approaches a constant value as t ® ¥.
c. Does velocity approach that constant value rapidly? Why or why not?
d. Derive that constant value of velocity without using the formula for v(t) found in part a.

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) = Ae^{k t}, where A is an arbitrary constant.

Solution

Let y = f(t). Then:



where A = ± e^c is an arbitrary non-0 constant. For A = 0, we have f(t) = 0, for which we get f '(t) = 0
= k . 0 = kf(t). So A = 0 also yields a solution. Thus, the solution is f(t) = Ae^{k t}, where A is an arbitrary
constant.

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