Calculus Of One
Real Variable


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1. FirstOrder Linear Differential Equations 
In this
section, the abbreviation “DE” stands for “differential equation”.
In Section
7.5 Part 3, we encountered the DE dy/dt = ky, or y' + by = 0, where b = – k. In this equation, the highest order of
derivative is 1 (the equation is of first order), and each term contains at
most one factor y or y', but not both, whose power is 1
(the equation is linear). It's called a firstorder linear differential
equation. Similarly, the equation x^{2}y  2x^{3}y'  e^{x} + 3 = 0 is a
firstorder linear DE. It can be converted to the form y'  (1/2x)y = (3  e^{x})/2x^{3} by changing the order of the first 2 terms,
moving the last 2 terms to the righthand side, and dividing the resulting
equation by 2x^{3}. This
form is called the standard
form of the DE.
A DE that can be written in the form: y' + p(x)y = q(x), where p(x) and q(x) are given continuous functions, is
called a firstorder linear differential equation. The form 
The highest
derivative order is 1, thus the adjective “firstorder”, and each term contains
at most either a factor y or a factor y'
raised to just the power of 1, thus the adjective “linear” (note: a term such
as y'y is of power 1 + 1 = 2). Remark that
the
equation is written in the decreasing order of the derivative (recall that y = y^{(0)}). The coefficient of y' is 1 and that of y is p(x).
Observe that while the equation is linear, each function y, y', p(x), and q(x) doesn't have to be linear. For
example, y = 2x + 3
is linear while y = x^{2} + 3 and q(x) = sin x  1 aren't.
The DE y' = f (x), as discussed in Section
16.1.1 Part 2, is a firstorder linear one. For this equation, p(x) = 0 and q(x) = f (x).
Suppose the
coefficient of y' of the DE:
f(x)y' + p_{1}(x)y = q_{1}(x)
is the
function f(x) that's not the constant function 1.
In this case, to get an equation of the form y' + p(x)y = q(x) that has 1
as the coefficient of y' and that's
equivalent to the given equation, we divide both sides of the given equation by
f(x):

Consider
the simple example of the particular case of the above DE, y' + by = 0, where p(x) = b and q(x) = 0, and
where b is
a constant. We solve it as follows:
y' + by = 0
y' = – by,
So y = Ce^{–}^{bx} + K isn't the general solution of y' + by = 0. The reason is that the general solution of a firstorder equation
involves only one arbitrary constant, and y = Ce^{–}^{bx} already has one arbitrary constant,
which is C. The arbitrary constant C
originates from the arbitrary constant C_{1}, which is added to bx to obtain the general antiderivative of b. This confirms that
we've already added an arbitrary constant.
Now let's
solve the DE by using an alternative approach. The lefthand side of y' + by = 0 has 2 terms, one term contains the
derivative y' as a
factor and the other term contains the function y as a factor. This reminds us of the product rule of
differentiation: (uv)' = u'v + uv', or (uv)' = uv' + u'v, where the
first term of the righthand side contains the derivative v'
as a factor and the other term contains the function v as a factor. So we try to examine
if y' + by is the derivative of the
product of some function, say u, and y, so that
the product is uy, whose
derivative is (uy)' = uy' + u'y. We see
that y' + by =
1y' + by, so it should be that u = 1 and u' = b. But b isn't the derivative of 1 unless b = 0. Thus y' + by, when b is non0,
in
itself can't be the derivative of the product of any function and y. The problem is that the
coefficient of y'
is 1. So we multiply
both sides of y'
+ by = 0 by a function u, obtaining uy'
+ buy = 0. In order for uy' + buy to be the derivative uy' + u'y of uy,
bu must be the derivative u' of u. We determine u so that this requirement is fulfilled, as follows:
u' = bu,
Multiplying
y' + by = 0 by e^{bx} works out well. The general solution is the same as obtained
previously.
We don't
need to memorize the formula of the general solution y of the DE y' + p(x)y = q(x). In solving such an equation,
we instead can follow the procedure that leads to that formula.
In the
general solution:
The
solution procedure of the firstorder linear DE can be summarized as follows.
Solution
Procedure Of The FirstOrder Linear Differential Equation
The firstorder linear DE of the standard form y' + p(x)y = q(x), where p(x) and q(x) are continuous, is solved by Make sure that the DE is in the standard form y' + p(x) y =
q(x),
where the coefficient of y’
is 1, because the above 
In Section
7.5 Part 3, we solved the DE dy/dt = ky, or y' + by = 0, where b = – k, and found the general solution to be y =
Ce^{kt}, where C = y(0). Use the integratingfactor technique described in this section to show
that the general solution of the DE:
y' + by = 0,
where y = y(x) is a
function of x and b is a constant, is y = y(0)e^{–}^{bx}. Note that we've dealt with this DE under the heading “A
Particular Case” above.
Here the
functions p(x) = b and q(x) = 0 are constant and thus
continuous.
Solution
An antiderivative of b is bx. Then:
e^{bx}y’ + e^{bx}by = 0 (multiply both sides of
the given DE by e^{bx}),
(e^{bx}y)’ = 0,
y(x) = y = Ce^{–}^{bx},
y(0) = Ce^{0} = C(1) = C,
y = y(0)e^{–}^{bx}.
EOS
Let's find
out what happens if we utilize the general antiderivative bx + c of b, where c is an arbitrary constant, instead
of the
simple particular antiderivative bx, where c = 0:
e^{bx}^{+}^{c}y’ + e^{bx}^{+}^{c}by = 0 (multiply both sides of the given DE by e^{bx}^{+}^{c}),
(e^{bx}^{+}^{c}y)’ = 0,
y(x) = y = Ce^{}^{bx}^{}^{c} = Ce^{}^{c}e^{}^{bx},
y(0) = Ce^{}^{c}e^{0} = Ce^{}^{c}(1) = Ce^{}^{c},
y = y(0)e^{–}^{bx}.
It's the
same general solution. Thus utilizing any particular antiderivative of p(x) is fine. We utilize the simplest one, where
the constant of integration is 0.
Solve y' + y/x = 2, where
x > 0.
Solution
EOS
Solve the
firstorder linear DE dy/dx + xy = 2x^{3}.
Solution
EOS
Example 1.4
Solve this
initialvalue problem:
Solution
EOS
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2. Applications 
We now
present examples of some of the applications of the firstorder linear DEs.
A tank
contains 200 L of brine in which 2.5 kg of salt is dissolved. Then brine
containing 0.1 kg of salt per liter enters the tank at
a rate of 15 L/min. While the brine is entering the tank, brine from the tank
is also exiting the tank at a rate of 8 L/min. The
concentration of the brine in the tank is kept uniform throughout the tank by
stirring. Find the amount of salt in the tank after 30
min.
Solution
Let S be the amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
enter and exit the tank. The rate of change of the amount of salt in the tank
at time t is:
C =
(17.5)(200^{8/7}),
After 30 min,
the amount of salt in the tank is approximately 33.3 kg.
EOS
We're to
determine a quantity. First we find the rate of change or derivative of that
quantity. We obtain a DE whose unknown
function is the quantity. Then we solve the DE and get the quantity.
A savings
account is opened with a deposit of $10,000. Money is being continuously
transferred automatically from another
account and deposited into this account in such a way that the accumulated
deposited amount increases at the rate of (1,000 +
200t) dollars
per year, where t is the
number of years after the account is opened. Interest is being paid also
continuously into
the account at the rate of 13% per year. Determine the balance of the account
after 5 years.
1. The quantity 1,000 + 200t is the rate of increase of the
accumulated deposited amount (ADA), not of the deposit, at time t.
For example, for the simpler case of
an ADA increasing at the constant rate of $5,000 per year, the deposit is
constant at
$5,000 per year and thus increases at
the rate of $0 per year.
2. Since the interest is paid
continuously, the accumulated interest (AI) at time t increases at (13%)B = $0.13B per year, where
B is the balance at time t. In this problem the unit of time
is the year. So the rate of increase of the AI (not the rate of
interest, not the rate of increase of
the rate of interest, the rate of interest being fixed at 13% per year) is also
$0.13B per
year.
3. Since the deposit is made
continuously (the number of deposits on any time interval approaches infinity),
we wonder if the
ADA is infinite and thus the balance
is infinite at any time t > 0 (the initial time t = 0 is when the account is opened). Let's
find out. As the rate of increase of
the ADA is 1,000 + 200t and the initial amount (at time t = 0) is $10,000, the ADA at
time t is a(t) = 1,000t + 200(t^{2}/2) + 10,000, or a(t) = 100t^{2} + 1,000t + 10,000. So at any time t where 0 < t < infinity, the
ADA a(t) is
finite. Although the parabola a(t) rises or
increases continuously for all increasing t > 0, its height at any t
where 0 < t < infinity is finite. For any
bounded interval [t_{1}, t_{2}] where 0 < t_{1} < t_{2}, divide
[t_{1}, t_{2}] into n equal subintervals,
and consider the parabola as
increasing n times on [t_{1}, t_{2}] from a(t_{1}) to a(t_{2}). For the continuous increase, let n approach
infinity. As the number of increases approaches
infinity, each increase approaches 0 and fast, fast enough for their sum to be
finite. Thus the height of the
parabola at any t where 0
< t <
infinity is finite. Although the deposits are made continuously,
each is small enough that the ADA is
finite. As for the interest, we already know that although it's paid
continuously, the
accumulated interest is finite. Thus
at any time t where 0
< t <
infinity the balance is finite.
Solution
Let B be the balance at time t. Then the rate of increase of the balance at time t is:
Let u = t and dv = e^{}^{0.13}^{t} dt, so that du = dt and v = e^{}^{0.13}^{t}/0.13. Then:
The balance
of the account after 5 years is $29,340.53.
EOS
Problems &
Solutions

1. Solve the DE y' – 2y/x = x^{2}.
An
antiderivative of –2/x is –2 ln x. Then:
2. Solve y' + y = 3e^{x}.
An
antiderivative of 1 is x, and so an integrating factor is e^{x}. We have:
Solution
y' + (cot x) y – 2x – 1 = 0,
y' + (cot x) y = 2x + 1.
4. Solve this initialvalue problem:
Solution
(1 + sin x) y' + (cos x) y = sin^{2} x,
5. A tank contains 240 L of brine in
which 6 kg of salt is dissolved. Then brine containing 0.06 kg of salt per
litre from outside
the tank flows into it at a rate of
24 L/min and the brine in it flows out of it at a rate of 12 L/min. The
concentration of brine
in the tank is kept uniform by
stirring. Determine the amount of salt in the tank after 10 minutes.
Solution
Let S be the
amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
flow into and out of the tank. The rate of change of the amount of salt in the
tank at time t is:
After 10
min, the amount of salt in the tank is 16 kg.
6. The initial balance of a savings
account was $2,000. Interest is paid into the account continuously at the
variable rate of (1 +
(t/60))% per month, where t is the number of months after the account was opened. Determine the
amount in the account
after 1 year.
Solution
Let a(t) be the amount in the account t months after the account was opened. The rate of increase of a(t) is:
2,000 = a(0) = Ce^{0} = C,
After 1
year or 12 months, the amount in the account is $2,282.22
7. The equation y' + p(x)y = q(x)y^{n}, where n is a constant different from 0 and 1, is known as Bernoulli equation.
(For n = 0,
the equation becomes y' + p(x)y = q(x), and for n = 1, it becomes y' + (p(x) – q(x))y = 0. In both
cases, the equation is
of firstorder and linear, and can be
solved by the integratingfactor technique discussed in this section.)
a. Prove that the substitution v = y^{1–}^{n} reduces the Bernoulli equation to a
DE for v that is of
firstorder and linear.
b. Use the result of part a to solve the DE y' – 2xy = 5xy^{3}.
a. Let v = y^{1–}^{n}. Then:
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