Math 217: Differential Equations
Lecture Notes: Substitution Methods - Homogeneous
First-Order DEs and Bernoulli Equations
Mark Pedigo
1
Substitution Methods
Substitution Method I: Homogeneous Equations
A homogeneous first-order DE is one that can be written in the form
dy
=F
dx
y
.
x
Note: Homogeneous has a very different meaning in the context of second-order
DEs. We’ll see this meaning in Chapter 2.
By an appropriate substitution, we can turn this equation into a separable equation. How
do we do this? First, we make the substitution v = xy . Then y = vx, so by the product rule
dy
dv
=v+x .
dx
dx
(1)
dv
= F (v) − v,
dx
(2)
Equation (1) is equivalent to
x
which is a separable equation.
Example 1.1. Solve the DE
dy
y+x
=
dx
x
Solution. We see
dy
y+x
=
dx
x
y
=
+ 1.
x
1
Plugging the substitutions
y = vx,
dy
dv
y
1
x
= v + x , v = , and = .
dx
dx
x
v
y
into the above equation yields
v+x
dv
dv
=v+1 ⇒ x
=1
dx
dx
1
⇒
dx = dv
x
dx
+C
x
⇒ v = ln |x| + C
⇒ v = ln |x| + ln C (changing C)
⇒ v = ln(|Cx|).
⇒
Since v = xy ,
y
x
Z
dv =
Z
= ln(Cx), or y = x ln(|Cx|).
Example 1.2. Solve the DE
2xy
dy
= 4x2 + 3y 2 .
dx
Solution. We see that
dy
4x2 + 3y 2
=
dx
2xy
2
4x
3y 2
=
+
2xy 2xy
!
x
3 y
= 2
+
.
y
2 x
Plugging the substitutions
y = vx,
dv
y
1
x
dy
= v + x , v = , and = .
dx
dx
x
v
y
into the above equation yields
dv
2 3
= + v.
dx
v 2
Subtracting v from both sides and finding a common denominator gives us
v+x
x
dv
2 v
=
+
dx
v 2
4 + v2
=
2v
v2 + 4
=
.
2v
2
(3)
(4)
(5)
Rearranging yields
2v
1
dv = dx.
+4
x
Now we integrate both sides and simplify.
v2
Z
Z
1
2v
dv
=
dx ⇒ ln(v 2 + 4) = ln x + ln C
2
v +4
x
⇒ v 2 + 4 = Cx.
This is our “solution” in terms of v. However, v was a substitution we introduced for
convenience. Changing v back to xy gives us
y2
+ 4 = Cx,
x2
or
y 2 + 4x2 = Cx3 .
Substitution Methods II: Bernoulli Equations
A first-order linear DE of the form
dy
+ P (x)y = Q(x)y n ,
dx
where is called a Bernoulli Equation. Note that if n = 0 or n = 1, this equation is a firstorder linear DE. Otherwise we transform the Bernoulli Equation into a linear first-order DE
as follows.
0
y + P (x)y = Q(x)y
n
y0
y
yn
⇒ n + P (x) n = Q(x) n
y
y
y
y0
1
⇒ n + P (x) · n−1 = Q(x).
y
y
0
(6)
(7)
1
Let v = yn−1
= y −(n−1) = y 1−n . Then v 0 = (1 − n)y −n y 0 = yyn · (1 − n). (We’re taking the
derivative of v and y with respect to x, so we used the chain rule.) Now, substituting v and
v 0 into equation (7) yields
v0
+ P (x)v = Q(x),
1−n
which is a linear first-order DE in v, which we know how to solve.
Example 1.3. Solve the DE y 0 =
y
x
− y2.
3
Solution. Dividing through by y 2 gives us
y0
1 1
· − 1.
=
y2
x y
Let v = y1 . Then v 0 = − y12 · y 0 . (Again, the differentiation is with respect to x, not y, so
we need the chain rule.) Substituting v yields
−v 0 =
R
v
1
− 1 ⇒ v 0 + v = 1.
x
x
1
Our integrating factor is e x dx = eln x = x. Multiplying through by this integrating
factor and solving the DE is as follows.
xv 0 + v = x ⇒ (xv)0 = x
⇒ xv =
Z
x dx + C
1
⇒ xv = x2 + C
2
x2 + C
x C
⇒ v= + =
2
x
2x
x2 + C
1
=
⇒
y
2x
2x
⇒ y= 2
x +C
4