The homogeneous first order constant coefficient linear differential equation is a differential equation of the
form
′
y + ay = 0. (23.1)
(23.1)
It is arguably the most important differential equation in existence. Generalizations of it include the entire
subject of linear differential equations and even many of the most important partial differential equations
occurring in applications.
Here is how to find the solutions to this equation. Multiply both sides of the equation by e^{at}. Then use
the product and chain rules to verify that
at ′ d-( at)
e (y + ay) = dt e y = 0.
Therefore, since the derivative of the function t → e^{at}y
(t)
equals zero, it follows this function
must equal some constant C. Consequently, ye^{at} = C and so y
(t)
= Ce^{−at}. This shows that if
there is a solution of the equation, y^{′} + ay = 0, then it must be of the form Ce^{−at} for some
constant, C. You should verify that every function of the form, y
(t)
= Ce^{−at} is a solution of the
above differential equation, showing that this yields all solutions. This proves the following
theorem.
Theorem 23.1.1The solutions to the equation, y^{′} + ay = 0 for a a real number consist of allfunctions of the form, Ce^{−at}where C is some constant.
Example 23.1.2Radioactive substances decay in the following way. The rate of decay is proportional tothe amount present. In other words, letting A
(t)
denote the amount of the radioactive substance at time t,A
(t)
satisfies the following initial value problem.
A′(t) = − k2A (t) ,A (0) = A
0
where A_{0}is the initial amount of the substance. What is the solution to the initial value problem?
and it only remains to find C. Letting t = 0, it follows A_{0} = A
(0)
= C. Thus A
(t)
= A_{0} exp
(− k2t)
.
Now here is another problem which is a little harder because it has something extra added in at the
end.
Example 23.1.3Find solutions to y^{′} = 2y + 1.
Here is how you do it:
Write as y^{′}− 2y = 1
Find an “Integrating Factor” ∫
(− 2)
dt = −2t. Note that I didn’t bother to add in the
arbitrary constant. This is because it does not matter. You don’t care about finding all
integrating factors. You just need one. Then an integrating factor is e^{−2t}.
Multiply both sides of the equation by the integrating factor.
e−2t(y′ − 2y) = d-(e−2ty (t)) = e−2t(1)
dt
Note that the first equal sign follows from the product rule and the chain rule. This is why wemultiply by the integrating factor, to get the derivative of something equal to somethingknown.
Take antiderivatives of both sides.
∫
e−2ty (t) = e−2tdt = − 1e−2t + C
2
Thus
1 2t
y (t) = − 2 + Ce
This time you need to be sure to keep the constant of integration because it does matter.
Note that by varying C you get different solutions to the differential equation. Now here are graphs of a
few of these solutions along with the slope field.
PICT
Note how the solutions follow the slope field. How do you determine the “right value” of C? This
involves an
An initial condition involves specifying a particular point which is to lie on the graph of the solution to
the differential equation. Then you can see from the picture that, having made this specification, the rest of
the graph should be determined by the need to follow the slope field. When you have specified the initial
condition as well as the differential equation, the problem is called an initial value problem.
Example 23.1.4Find the solution to the initial value problem
y′ = 2y +1, y(1) = 2
From the above example, all solutions are of the form y = −
1
2
+ Ce^{2t}. It is now just a matter of finding
the value of C which will cause the given point
(1,2)
, expressed by saying that y
(1)
= 2,
to lie on the graph of y. Thus you need to have 2 = −
1
2
+ Ce^{2}. Then you just need to solve
this equation for C. This yields C =
5--
2e2
. Therefore, thesolution to the initial value problem
is
y = − 1 +-5-e2t
2 2e2
Note the use of the definite article. There is only one solution to this initial value problem although there
are infinitely many solutions to the differential equation, three of which were graphed above. This
uniqueness property will be discussed more later, but for now, you can see roughly why this is. It comes
from the need for the solution to follow the slope field, so if you specify a point on the curve, you have
essentially determined it.
Example 23.1.5Find the solution to the initial value problem
∫ t
y(t) = exp(− A (t))exp(A (t0))y(t0)+ exp (− A (t)) exp(A (s))b(s)ds
∫ t t0
= exp(A(t0)− A (t))y0 + exp(A(s)− A (t))b(s)ds
t0
This shows that if the linear initial value problem has a solution, then it must be of the
above form. Hence there is at most one solution to the initial value problem. Does the above
formula actually give a solution to the initial value problem? Let y
so the initial condition holds. Does it solve the differential equation? By the chain rule and the fundamental
theorem of calculus,
′ ′
y (t) = (− A (t))exp(A (t0)− A∫(t))y0 + exp(− A(t))exp (A(t)) b(t)
+ (− A ′(t))exp(− A (t)) texp(A (s))b(s)ds
t0
= (− a(t))exp (A(t0)− A(t))y0 + exp(− A (t))exp(A (t))b(t)
∫ t
+ (− a(t))exp (− A (t)) exp(A(s))b(s)ds = − a(t)y (t)+ b(t)
t0
so it also is a solution of the linear initial value problem.
Example 23.1.10This example illustrates a different notation for differential equations. Find thesolutions to
xdy + (2xy− xsinx)dx = 0
The idea is you divide by dx and so the exact meaning is
xy′ + 2xy = xsin(x)
Then
y′ + 2y = sin x, (e2xy)′ = e2x sinx
∫ 1
e2xy = e2x sin(x)dx = 5e2x(2sin x− cosx)+ C
1
y = -(2sin x− cosx)+ Ce −2x
5
The reason for writing it this way is that sometimes you want to find x as a function of y and this notation
is neutral in terms of which variable is the independent variable.
Example 23.1.11A radioactive substance decays in such a way that the rate of change of theamount of the substance is a constant multiple of the amount present, the constant being negative.Thus
dAdt
= −kA. There is a certain sample of decaying material. Measurements are taken after 5years and it is found that there is about 9∕10 of the original amount present. Find the half life ofthis material. The half life is the amount of time it takes for half of it to have decayed.
From the equation, A = A_{0}e^{−kt}. Then
910
A_{0} = A_{0}e^{−k(5)
}. Solving this for k yields
− ln(.9)
--5---
= k and so the
amount of time to have half of what was started with is T given as a solution to the following
equation.
− ln(.9) 1 ln(.5)
e−( 5 )(T) = -, so T =--------= 32.894
2 ln (.9)∕5
This kind of thing is associated not just with radioactive material but with other chemicals as well.
They degrade over time according to such an equation.
Example 23.1.12The ancient Babylonians were fascinated with the idea of compound interest. Theywere interested in how long it would take an initial amount to double. One can understand compoundinterest compounded continuously using the same kind of differential equation as the above only this timethe constant is positive and is the interest rate. Thus
dA- = kA
dt
If the interest rate is 20% per year compounded continuously, how long will it take for an initial amount todouble in size?
From the equation, A = A_{0}e^{.2t} where A_{0} is the initial amount. Then you want to find T such that
2A_{0} = A_{0}e^{.2T} and so
ln-2
T = .2 = 5.0ln2 = 3.4657
If the rate is r per year and you have n years and the interest is compounded at the end of each year
rather than continuously, then the amount is given by the formula
(1+ r)
^{n} = A
(n)
. Anciently, they used
this kind of thing because they did not have differential equations. If the interest rate is 20% compounded
monthly, then the amount after n years is A_{0}
( )
1+ 1.22
^{12n}where A_{0} is the initial amount. If n = 3.5, a use of
a calculator shows that
( -.2 )12(3.5)
1+ 12 = 2.0022
which is very similar to compounding the interest continuously. The rational for this formula is that if it is
compounded monthly, then the interest rate per month is .2∕12. Each successive month is called a payment
period.
Example 23.1.13A lake contains one million gallons of water. A gas tank starts to leak upstreamand contaminated water mixed with gasoline starts flowing into the lake at the rate of 1000 gallonsper month.This is mixed well due to large numbers of fish in the lake and water flows out at thesame rate. The amount of gasoline in the contaminated water varies due to the demand for gas atthe gas station and the concentration of gasoline in the contaminated water is
(1 + sin(t))
gramsper gallon. Find a formula for the concentration of gasoline in the lake in grams per gallon as afunction of time in months after a long time.
Let A be the amount of gas in the lake. Then
dA A 1
-dt = (1 + sin(t)) × 1000 − 1061000 = 1000(1+ sin(t))− 103-A
Rather than worry with the stupid numbers, write this as
A ′ + aA = b(1 + sin(t)), A (0) = 0
Following the procedure for finding solutions to a linear equation,
(eatA )′ = b(1+ sin(t))eat
Now it follows that, taking antiderivatives of both sides,
( )
at --eat--( 2 2 )
e A = b a3 + a a sin t− acost+ a + 1 + C
Since A
(0)
= 0, it follows that
( )
0 = b --1---(− a+ a2 +1) + C
a3 + a
and so C = −
a3b+a
( )
a2 − a + 1
. Therefore,
( ) ( 2 ) −at
A = be−at -eat--(a2 sint− a cost + a2 + 1) +-a-−-a-−-1-be--
a3 + a a3 + a
Now placing in the formula the values of a and b and then simplifying the result it follows that A
equals
After a long time, the terms having the negative exponential will disappear in the limit and this yields for
the number of grams per gallon the formula
1− 0.001cost+ 1.0 × 10− 6sint
Note that this yields approximately 1 gram per gallon. Compare to the concentration of the incoming
water. The concentration of the incoming water oscillates about 1 and so does the concentration of gas in
the lake, although the oscillations are much much smaller. This is due to the large number of gallons in the
lake. You might have expected this but you could not have predicted exact values without the differential
equation.
Example 23.1.14A pumpkin is launched 30^{∘}from the horizontal at a speed of 60 feet per second.It is acted on by the force of gravitywhich delivers an acceleration which is 32 feet per second squaredand an acceleration due to air resistance which we assume is .2 times the speed which acts in theopposite direction to the direction of motion. Describe the position of the pumpkin as a function oftime.