4First-order differential equations
IA Differential Equations
4.5 Non-linear equations
In general, a first-order equation has the form
Q(x, y)
dy
dx
+ P (x, y) = 0.
(this is not exactly the most general form, since theoretically we can have powers
of
y
0
or even other complicated functions of
y
0
, but we will not consider that
case here).
4.5.1 Separable equations
Definition (Separable equation). A first-order differential equation is separable
if it can be manipulated into the following form:
q(y) dy = p(x) dx.
in which case the solution can be found by integration
Z
q(y) dy =
Z
p(x) dx.
This is the easy kind.
Example.
(x
2
y − 3y)
dy
dx
− 2xy
2
= 4x
dy
dx
=
4x + 2xy
2
x
2
y − 3y
=
2x(2 + y
2
)
y(x
2
− 3)
y
2 + y
2
dy =
2x
x
2
− 3
dx
Z
y
2 + y
2
dy =
Z
2x
x
2
− 3
dx
1
2
ln(2 + y
2
) = ln(x
2
− 3) + C
ln
p
2 + y
2
= ln A(x
2
− 3)
p
y
2
+ 2 = A(x
2
− 3)
4.5.2 Exact equations
Definition (Exact equation).
Q
(
x, y
)
dy
dx
+
P
(
x, y
) = 0 is an exact equation iff
the differential form
Q
(
x, y
) d
y
+
P
(
x, y
) d
x
is exact, i.e. there exists a function
f(x, y) for which
df = Q(x, y) dy + P (x, y) dx
If
P
(
x, y
) d
x
+
Q
(
x, y
) d
y
is an exact differential of
f
, then d
f
=
P
(
x, y
) d
x
+
Q
(
x, y
) d
y
. But by the chain rule, d
f
=
∂f
∂x
d
x
+
∂f
∂y
d
y
and this equality holds
for any displacements dx, dy. So
∂f
∂x
= P,
∂f
∂y
= Q.
From this we have
∂
2
f
∂y∂x
=
∂P
∂y
,
∂
2
f
∂x∂y
=
∂Q
∂x
.
We know that the two mixed 2nd derivatives are equal. So
∂P
∂y
=
∂Q
∂x
.
The converse is not necessarily true. Even if this equation holds, the differential
need not be exact. However, it is true if the domain is simply-connected.
Definition (Simply-connected domain). A domain
D
is simply-connected if it
is connected and any closed curve in
D
can be shrunk to a point in
D
without
leaving D.
Example. A disc in 2D is simply-connected. A disc with a “hole” in the middle
is not simply-connected because a loop around the hole cannot be shrunk into a
point. Similarly, a sphere in 3D is simply-connected but a torus is not.
Theorem. If
∂P
∂y
=
∂Q
∂x
through a simply-connected domain
D
, then
P
d
x
+
Q
d
y
is an exact differential of a single-valued function in D.
If the equation is exact, then the solution is simply
f
= constant, and we
can find f by integrating
∂f
∂x
= P and
∂f
∂y
= Q.
Example.
6y(y − x)
dy
dx
+ (2x − 3y
2
) = 0.
We have
P = 2x − 3y
2
, Q = 6y(y − x).
Then
∂P
∂y
=
∂Q
∂x
= −6y. So the differential form is exact. We now have
∂f
∂x
= 2x − 3y
2
,
∂f
∂y
= 6y
2
− 6xy.
Integrating the first equation, we have
f = x
2
− 3xy
2
+ h(y).
Note that since it was a partial derivative w.r.t.
x
holding
y
constant, the
“constant” term can be any function of
y
. Differentiating the derived
f
w.r.t
y
,
we have
∂f
∂y
= −6xy + h
0
(y).
Thus h
0
(y) = 6y
2
and h(y) = 2y
3
+ C, and
f = x
2
− 3xy
2
+ 2y
3
+ C.
Since the original equation was d
f
= 0, we have
f
= constant. Thus the final
solution is
x
2
− 3xy
2
+ 2y
3
= C.