Differential Equations: MTH 2140 Lecture 2
1
Solutions and Integral Curves
Definition 1. Consider
du(t)
= −k(u(t) − T )
dt
(1)
By a solution of a differential equation (1), we mean a differentiable function u = ϕ(t),
which satisfies a differential equation. A specific instance of a solution to equation (2) is
represented by u = T0 . This is evident as the substitution of u = T0 into the equation
results in the simplification of (2) to the identity 0 = 0. The constant solution u = T0 is
recognized as an equilibrium solution for the differential equation (2). These equilibrium
solutions typically hold significant importance in comprehending the behavior exhibited by
other solutions.
Let us suppose that the ambient temperature T in (1) is a constant, say, T = T0 , so that
(1) becomes
du(t)
= −k(u(t) − T0 )
dt
(2)
If we assume that u ̸= T0 in (2), we can discover other solutions by first rewriting it in the
form
1 du
= −k.
u − T0 dt
Multiplying by dt, we get
1
du = −k dt.
u − T0
Then, by integrating both sides of the above equation, we obtain
ln |u − T0 | = −kt + C,
where C is an arbitrary constant of integration. Next, we solve for u. Taking the exponential
of both sides, we have
|u − T0 | = e−kt+C = eC e−kt ,
or
u − T0 = ±eC e−kt .
1
Thus,
u = T0 + ce−kt ,
(3)
is a solution to (2), where c = ±eC is also an arbitrary nonzero constant. If c = 0, then we
obtain an equilibrium solution. The expression (3) contains all the possible solutions of (2)
and is known as the general solution of the equation.
If we were asked to verify that a given function is a solution of a given equation, then
we would need to substitute the given function in the given equation and show that the
equation reduces to an identity. For example,
Example 1. Show that ϕ(x) = x2 − x−1 is a solution to
x2
d2 y
= 2y,
dx2
but φ = x3 is not.
2
(4)
1.0.1
Integral Curve
The geometrical representation of the general solution (3) is an infinite family of curves in
the tu− plane called integral curves. Each integral curve is associated with a particular value
of c; it is the graph of the solution corresponding to the value of c.
With k = 1.5 day −1 and T0 = 60◦ F in (2) and (3) gives us
du
= −1.5(u − 60)
dt
(5)
and the corresponding general solution becomes
u = 60 + ce−1.5t .
(6)
The following figure shows several integral curves of (5) obtained by plotting the graphs of
the function in (6) for different values of c. Note that all solutions approach the equilibrium
solution u = 60 as t → ∞.
3
1.0.2
Initial Value Problems
We know from (3), that a differential equation has infinitely many solutions. In order to
establish a unique solution, it becomes necessary to introduce supplementary prerequisites
recognized as initial conditions.
Definition 2. By an initial value problem (IVP) for an nth-order differential equation
F (t, y, y ′ , y ′′ , · · · , y n ) = 0,
(7)
we mean: Find a solution to the differential equation on an interval I that satisfies at t0 the
n initial conditions
y(t0 ) = y0 ,
y ′ (t0 ) = y1 ,
···
···
···
y n−1 (t0 ) = yn−1 ,
where t0 ∈ I and y0 , y1 , · · · yn−1 are given constants.
In the case of a first-order differential equation, the initial conditions reduce to the single
requirement
y(t0 ) = y0 ,
and in the case of a second-order differential equation, the initial conditions have the form
y ′ (t0 ) = y1 .
y(t0 ) = y0 ,
The phrase “initial conditions” originates from mechanics and uses the independent variable
t to denote time. If t0 is the starting time, y(t0 ) = y0 represents the initial location of an
object, and y ′ (t0 ) = y1 gives its initial velocity. If y ” = 0 say that the graph of y(t) is a line,
so it is uniquely determined by specifying a point on the line.
y(t0 ) = y0 ,
and the slope of the line,
y ′ (t0 ) = y1 .
Example 2. Show that ϕ(x) = sinx − cosx is a solution to the initial value problem
′′
y + y = 0;
y(0) = −1,
4
y ′ (0) = 1.