23.1 First Order Linear Equations
The homogeneous first order constant coefficient linear differential equation is a differential equation of the
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 eat. Then use
the product and chain rules to verify that
Therefore, since the derivative of the function t → eaty
equals zero, it follows this function
must equal some constant
and so y
This shows that if
there is a solution of the equation, y′
then it must be of the form Ce−at
You should verify that every function of the form, y
is a solution of the
above differential equation, showing that this yields all solutions. This proves the following
Theorem 23.1.1 The solutions to the equation, y′ + ay = 0 for a a real number consist of all
functions of the form, Ce−at where C is some constant.
Example 23.1.2 Radioactive substances decay in the following way. The rate of decay is proportional to
the amount present. In other words, letting A
denote the amount of the radioactive substance at time t,
satisfies the following initial value problem.
where A0 is the initial amount of the substance. What is the solution to the initial value problem?
Write the differential equation as A′
From Theorem 23.1.1
and it only remains to find C. Letting t = 0, it follows A0 = A
Now here is another problem which is a little harder because it has something extra added in at the
Example 23.1.3 Find solutions to y′ = 2y + 1.
Here is how you do it:
- Write as y′− 2y = 1
- Find an “Integrating Factor” ∫
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.
Note that the first equal sign follows from the product rule and the chain rule. This is why we
multiply by the integrating factor, to get the derivative of something equal to something
- Take antiderivatives of both sides.
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.
Note how the solutions follow the slope field. How do you determine the “right value” of C? This
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 23.1.4 Find the solution to the initial value problem
From the above example, all solutions are of the form y = −
It is now just a matter of finding
the value of C
which will cause the given point
expressed by saying that y
to lie on the graph of y
. Thus you need to have 2 = −
Then you just need to solve
this equation for C
. This yields C
solution to the initial value problem
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 23.1.5 Find the solution to the initial value problem
- Find the integrating factor. ∫
2tdt = t2. Integrating factor: exp =
- Multiply both sides by the integrating factor.
- Take ∫
of both sides.
- Solve for y
- Find C to satisfy the initial condition.
- Place value of C you just found in the formula for y
Now at this point, you should check and see if it works. It needs to solve both the initial condition and
the differential equation.
Example 23.1.6 Find the solutions to
- Find integrating factor. A +
. Integrating factor: exp
- Multiply by exp
- Do ∫
to both sides. Pick F
This proves the following theorem.
Theorem 23.1.7 The solutions to the equation, y′ + a
consist of all functions of the
C where F
dt and C is a constant, A′
Finally, here is a uniqueness theorem.
Theorem 23.1.8 If a
is a continuous function, there is at most one solution to the initial value
Proof: If there were two solutions y1 and y2, then letting w = y1 −y2, it follows w′ + a
= 0 and
Then multiplying both sides of the differential equation by eA
= 0 and so eA
for some constant, C.
= 0 and so this constant can
only be 0. Hence
= 0 and so y1
Finally, consider the general linear initial value problem.
Definition 23.1.9 A linear differential equation is one which is of the form
where a,b are continuous. The corresponding initial value problem is
Now here are the steps for solving the initial value problem.
- Find the integrating factor ∫
dt ≡ A +
C. The integrating factor is exp =
- Multiply both sides by the integrating factor.
Why is this so? It involves the chain rule and the product rule.
- Next do ∫
t0t to both sides.
Then by the fundamental theorem of calculus,
and so, you can solve for y and get
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
be given by that formula.
so the initial condition holds. Does it solve the differential equation? By the chain rule and the fundamental
theorem of calculus,
so it also is a solution of the linear initial value problem.
Example 23.1.10 This example illustrates a different notation for differential equations. Find the
The idea is you divide by dx and so the exact meaning is
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 23.1.11 A radioactive substance decays in such a way that the rate of change of the
amount of the substance is a constant multiple of the amount present, the constant being negative.
−kA. There is a certain sample of decaying material. Measurements are taken after 5
years and it is found that there is about
10 of the original amount present. Find the half life of
this material. The half life is the amount of time it takes for half of it to have decayed.
From the equation, A = A0e−kt. Then
Solving this for k
and so the
amount of time to have half of what was started with is T
given as a solution to the following
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 23.1.12 The ancient Babylonians were fascinated with the idea of compound interest. They
were interested in how long it would take an initial amount to double. One can understand compound
interest compounded continuously using the same kind of differential equation as the above only this time
the constant is positive and is the interest rate. Thus
If the interest rate is 20% per year compounded continuously, how long will it take for an initial amount to
double in size?
From the equation, A = A0e.2t where A0 is the initial amount. Then you want to find T such that
2A0 = A0e.2T and so
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
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 A0
is the initial amount. If n
a use of
a calculator shows that
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
Example 23.1.13 A lake contains one million gallons of water. A gas tank starts to leak upstream
and contaminated water mixed with gasoline starts flowing into the lake at the rate of 1000 gallons
per month. This is mixed well due to large numbers of fish in the lake and water flows out at the
same rate. The amount of gasoline in the contaminated water varies due to the demand for gas at
the gas station and the concentration of gasoline in the contaminated water is
per gallon. Find a formula for the concentration of gasoline in the lake in grams per gallon as a
function of time in months after a long time.
Let A be the amount of gas in the lake. Then
Rather than worry with the stupid numbers, write this as
Following the procedure for finding solutions to a linear equation,
Now it follows that, taking antiderivatives of both sides,
it follows that
and so C = −
Now placing in the formula the values of a and b and then simplifying the result it follows that A
Then, dividing by the number of gallons in the lake, this yields for the number of grams per
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
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
Example 23.1.14 A pumpkin is launched 30∘ from the horizontal at a speed of 60 feet per second.
It is acted on by the force of gravity which delivers an acceleration which is 32 feet per second squared
and an acceleration due to air resistance which we assume is .2 times the speed which acts in the
opposite direction to the direction of motion. Describe the position of the pumpkin as a function of
Let the initial position be at
and let the coordinates of the point be
What is the
initial velocity? It is
Then the acceleration is given by
Thus y′′ + .2y′ = −32. Let’s solve for y′.
So what is C? When t = 0, we get C − 160 = 30 and so C = 190. Hence
What is D? When t = 0 we want y
= 0 and so