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Note: 1 lecture (or less), §1.2 in [EP], covered in §1.2 and §2.1 in [BD]

A ﬁrst order ODE is an equation of the form

or just

In general, there is no simple formula or procedure one can follow to ﬁnd solutions. In the next few lectures we will look at special cases where solutions are not diﬃcult to obtain. In this section, let us assume that is a function of alone, that is, the equation is

(1.1) |

We could just integrate (antidiﬀerentiate) both sides with respect to .

that is

This is actually the general solution. So to solve (1.1), we ﬁnd some antiderivative of and then we add an arbitrary constant to get the general solution.

Now is a good time to discuss a point about calculus notation and terminology. Calculus textbooks muddy the waters by talking about the integral as primarily the so-called indeﬁnite integral. The indeﬁnite integral is really the antiderivative (in fact the whole one-parameter family of antiderivatives). There really exists only one integral and that is the deﬁnite integral. The only reason for the indeﬁnite integral notation is that we can always write an antiderivative as a (deﬁnite) integral. That is, by the fundamental theorem of calculus we can always write as

Hence the terminology to integrate when we may really mean to antidiﬀerentiate. Integration is just one way to compute the antiderivative (and it is a way that always works, see the following examples). Integration is deﬁned as the area under the graph, it only happens to also compute antiderivatives. For sake of consistency, we will keep using the indeﬁnite integral notation when we want an antiderivative, and you should always think of the deﬁnite integral as a way to write it.

Example 1.1.1: Find the general solution of .

Elementary calculus tells us that the general solution must be . Let us check by diﬀerentiating: . We got precisely our equation back.

Normally, we also have an initial condition such as for some two numbers and ( is usually 0, but not always). We can then write the solution as a deﬁnite integral in a nice way. Suppose our problem is , . Then the solution is

(1.2) |

Let us check! We compute , via the fundamental theorem of calculus, and by Jupiter, is a solution. Is it the one satisfying the initial condition? Well, . It is!

Do note that the deﬁnite integral and the indeﬁnite integral (antidiﬀerentiation) are completely diﬀerent beasts. The deﬁnite integral always evaluates to a number. Therefore, (1.2) is a formula we can plug into the calculator or a computer, and it will be happy to calculate speciﬁc values for us. We will easily be able to plot the solution and work with it just like with any other function. It is not so crucial to always ﬁnd a closed form for the antiderivative.

By the preceding discussion, the solution must be

Here is a good way to make fun of your friends taking second semester calculus. Tell them to ﬁnd the closed form solution. Ha ha ha (bad math joke). It is not possible (in closed form). There is absolutely nothing wrong with writing the solution as a deﬁnite integral. This particular integral is in fact very important in statistics.

Using this method, we can also solve equations of the form

Let us write the equation in Leibniz notation.

Now we use the inverse function theorem from calculus to switch the roles of and to obtain

What we are doing seems like algebra with and . It is tempting to just do algebra with and as if they were numbers. And in this case it does work. Be careful, however, as this sort of hand-waving calculation can lead to trouble, especially when more than one independent variable is involved. At this point we can simply integrate,

Finally, we try to solve for .

Example 1.1.3: Previously, we guessed (for some ) has the solution . We can now ﬁnd the solution without guessing. First we note that is a solution. Henceforth, we assume . We write

We integrate to obtain

where is an arbitrary constant. Now we solve for (actually for ).

If we replace with an arbitrary constant we can get rid of the absolute value bars (which we can do as was arbitrary). In this way, we also incorporate the solution . We get the same general solution as we guessed before, .

Example 1.1.4: Find the general solution of .

First we note that is a solution. We can now assume that . Write

We integrate to get

We solve for . So the general solution is

Note the singularities of the solution. If for example , then the solution “blows up” as we approach . Generally, it is hard to tell from just looking at the equation itself how the solution is going to behave. The equation is very nice and deﬁned everywhere, but the solution is only deﬁned on some interval or .

Classical problems leading to diﬀerential equations solvable by integration are problems dealing with velocity, acceleration and distance. You have surely seen these problems before in your calculus class.

Example 1.1.5: Suppose a car drives at a speed meters per second, where is time in seconds. How far did the car get in 2 seconds (starting at )? How far in 10 seconds?

Let denote the distance the car traveled. The equation is

We just integrate this equation to get that

We still need to ﬁgure out . We know that when , then . That is, . So

Thus and

Now we just plug in to get where the car is at 2 and at 10 seconds. We obtain

Example 1.1.6: Suppose that the car accelerates at a rate of . At time the car is at the 1 meter mark and is traveling at 10 ^{m}/_{s}. Where is the car at time .

Well this is actually a second order problem. If is the distance traveled, then is the velocity, and is the acceleration. The equation with initial conditions is

What if we say . Then we have the problem

Once we solve for , we can integrate and ﬁnd .

Exercise 1.1.9: A spaceship is traveling at the speed ^{km}/_{s} ( is time in seconds). It is pointing directly away from earth and at time it is 1000 kilometers from earth. How far from earth is it at one minute from time ?

Exercise 1.1.11: A dropped ball accelerates downwards at a constant rate meters per second squared. Set up the diﬀerential equation for the height above ground in meters. Then supposing meters, how long does it take for the ball to hit the ground.

Exercise 1.1.104: Sid is in a car traveling at speed miles per hour away from Las Vegas, where is in hours. At , Sid is 10 miles away from Vegas. How far from Vegas is Sid 2 hours later?

Exercise 1.1.106: The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius centimeters is . The surface area is . Set up the diﬀerential equation for how is changing. Then, suppose that at time minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time will the snowball be completely melted.