Classification and Linear Equations

2.2. Explicit Equations. The simplest form of (2.1) to treat is that of so-called explicit
equations,

In this case the deriviative is given as an explicit function of t. This case is usually covered
in calculus courses, so we only review it here.

2.2.1. Recipe for Solving Explicit Equations. You should recall that a differentiable func -
tion F is said to be a primitive or antiderivative of f if F′ = f. You thereby see that
y = Y (t) will be a solution of (2.3) if and only if Y is a primitive of f. You should also
recall that if you know one primitive F of f then any other primitive Y of f must have the
form Y (t) = F(t) + c for some constant c. We thereby see that if (2.3) has one solution
then it has a family of solutions given by the indefinite integral of f — namely, by

,where F′(t) = f(t) and c is any constant . (2.4)

Moreover, there are no other solutions of (2.3). The family (2.4) is therefore called a
general solution of the differential equation (2.3).

2.2.2. Initial- Value Problems for Explicit Equations. In order to pick a unique solution
from the family (2.3) one must impose an additional condition that determines c. We do
this by imposing a so-called initial condition of the form

where is called the initial time and is called the initial value or initial datum. The
combination of the differential equation (2.2) with the above initial condition is

This is a so-called an initial-value problem. By imposing the initial condition upon the
family (2.4) we see that

which implies that . Therefore, if f has a primitive F then the unique
solution of initial-value problem (2.5) is given by

The above arguments show that the problems of finding either a general solution of
(2.3) or the unique solution of the initial-value problem (2.5) reduce to the problem of
finding a primitive F of f. Given such an F, a general solution of (2.3) is given by (2.4)
while the unique solition of initial-value problem (2.5) is given by (2.6). These arguments
however do not insure that such a primitive exists. Of course, for sufficiently simple f you
can find a primitive analytically.

Example: Find a general solution to the differential equation

Solution: By (2.4) a general solution is

Example: Find a solution to the initial-value problem

Solution: The previous example shows the solution has the form w = 2x³+x+c for some
constant c. Imposing the initial condition gives 2 · 1³ + 1 + c = 5, which implies c = 2.
Hence, the solution is w = 2x³ + x + 2.

Alternative Solution: By (2.6) with , and the primitive F(x) = 2x³ + x
we find

Remark. As the solutions to previous example illustrate, when solving an initial-value
problem it is often easier to first find a general solution and then evaluate the c from the
initial condition rather than to directly apply formula (2.6). With that approach you do
not have to memorize formula (2.6).

2.2.3. Existence of Solutions for Explicit Equations. Finally, even when you cannot find
a primitive analytically, you can show that a solution exists by appealing to the Second
Fundamental Theorem of Calculus. It states that if f is continuous over an interval (a, b)
then for every to in (a, b) one has

In other words, f has a primitive over (a, b) that can be expressed as a definite integral.
Here s is the “dummy” variable of integ ration in the above definite integral. If is in
(a, b) then the First Fundamental Theorem of Calculus implies that formula (2.6) can be
expressed as

This shows that if f is continuous over an interval (a, b) that contains then the initial-
value problem (2.5) has a unique solution over (a, b), which is given by formula (2.7). This
formula can be approximated by numerical quadrature for any such f.

Definition: The largest interval over which a solution exists is called its interval
of existence or interval of definition.

For explicit equations one can usually identify the interval of existence for the solution of
the initial-value problem (2.5) by simply looking at f(t). Specifically, if Y (t) is the solution
of the initial value problem (2.5) then its interval of existence will be whenever:

• f(t) is continuous over ,
• the initial time is in ,
• f(t) is not defined at both and .

This is because the first two bullets along with the formula (2.7) imply that the interval
of existence will be at least , while the last two bullets along with our definition
(2.2) of solution imply that the interval of existence can be no bigger than . This
argument works when or .

2.3. Linear Equations . The next simplest form of (2.1) to treat is that of so-called
linear equations,

In this case the derivative of y is given as a linear function of y whose coefficients are
functions of t. It contains the explicit case when a(t) = 0.

More generally, every linear first-order ODE for a single unknown function y(t) can
be brought into the form

where p(t), q(t), and r(t) are given functions of t such that p(t) ≠ 0 for those t over which
the equation is considered. The functions p(t) and q(t) are called coefficients while the
function r(t) is called the forcing or driving.

A linear equation may not be given to you in the form (2.8). However, you can put it
into this form by simply grouping all the terms involving either the unkown function or its
derivative on the left-hand side, while grouping all the other terms on the right-hand side.

Example. Consider the equation

You should be able to see that this equation is linear and to bring it into the form (2.8).

Solution. By grouping all the terms involving either the z or its derivative on the left-hand
side, while grouping all the other terms on the right-hand side, we obtain

This is in the form (2.8).

2.3.1. Recipe for Solving Linear Equations. The following is a straightforward recipe
that reduces the problem of generating an analytic solution of (2.8) to that of finding two
primitives. One first brings (2.8) into the normal form by dividing by p (t). This yields

Below we will show that this is equivalent to the so -called integrating factor form

This is an explicit equation for the derivative of that can be integrated to obtain
where and c is any constant .

(2.11)

A general solution of (2.8) is therefore given by the family

If you are solving an initial-value problem then you can evaluate c from the initial condition.

The key to understanding the above recipe is to understand the equivalence of the
normal form (2.9) and integrating factor form (2.10). This equivalence follows from the
fact that

This calculation shows that equation (2.10) is simply equation (2.9) multiplied by .
Because the factor is always positive, the equations are therefore equivalent. We call
an integrating factor of equation (2.9) because after multiplying both sides of (2.9)
by the left-hand side can be written as the derivative of An integrating factor
thereby allows you to reduce the linear case to the explicit case.

Rather than memorizing formula (2.12), it is easier to approach first-order linear
ordinary differential equations by simply retracing the steps by which (2.12) was derived.
We illustrate this approach with the following examples.

Example. Find the general solution to

Solution. First bring the equation into the normal form

An integrating factor is where A′(t) = 5. Setting A(t) = 5t, we then bring the
equation into the integrating factor form

By integrating both sides of this equation we obtain

The general solution is therefore given by