INTEGRATING FACTORS
5 minute review. Remind students how to use integrating factors to solve linear
dy
first-order differential equations. The example dx
+2xy = x is a good one, if needed.
Class warm-up. Find the solution to
dy
8
+ 8x7 y = ex−x ,
dx
satisfying y = 2 when x = 1.
Problems. Choose from the below.
1. A basic example. Find the general solution to the differential equation
dy 2y
sin x
+
= 2 .
dx
x
x
2. An example with initial conditions. Find the solution of the differential
equation
dy
y
exp
+
=x
dx x
which satisfies x = 1 at y = 1.
3. Daily temperature cycles. Recall Newton’s law of cooling, which gives a
differential equation describing the temperature of a body T in terms of the
ambient temperature Ta , namely
dT
= −k(T − Ta ).
dt
In the videos we solved this differential equation when Ta was constant. Instead, let’s assume Ta = sin t (a crude approximation of a daily temperature
cycle). Find the solution of the resulting differential equation.
4. A non-linear equation. By considering the expansion of (s − t)(s + t − u),
find two independent families of solutions to
2
dy
dy
−x
− y 2 + xy = 0.
dx
dx
1
2
INTEGRATING FACTORS
8
For the warm-up, the integrating factor is ex , and the particular solution is y =
8
8
ex−x + e1−x .
Selected answers and hints.
1. It’s
y=
c − cos x
.
x2
2. The solution turns out to be
1
1
5
y = x ln x − x +
.
2
4
4x
3. The integrating factor is (as in the example considered in the videos) ekt , and
the solutions turn out to be
k(k sin t − cos t)
+ ce−kt ,
T =
k2 + 1
where c is a constant.
dy
dy
4. The equation factorises as dx
= 0, and we can find one
−y
+
y
−
x
dx
family of solutions to each bracket. This leads to the solutions y = Aex and
y = x − 1 + Be−x .
For more details, start a thread on the discussion board.