SUBSTITUTION
MINGFENG ZHAO
September 15, 2014
SUBSTITUTION
Example 1. Solve y 0 = (x − y + 1)2 .
Let v = x − y + 1, then
v0 = 1 − y0 .
That is,
y0 = 1 − v0 = v2 .
So v 0 = 1 − v 2 . Let’s find the general solution to v 0 = 1 − v 2 :
1) v(x) ≡ ±1 are solutions. Since v = x − y + 1, then
either y = x,
or y = x + 2.
2) If v(x) 6≡ ±1, then
Z
x=
1 v + 1 1
dv
=
ln
+ C.
1 − v2
2 v − 1
Then
v+1
= ±e2(x−C) = ±e−2C · e2x .
v−1
Since C is arbitrary constant, then
v+1
= Ce2x ,
v−1
for some nonzero constant C.
Then
v + 1 = Ce2x v − Ce2x .
That is,
v=
Ce2x + 1
.
Ce2x − 1
1
2
MINGFENG ZHAO
Since v = x − y + 1, then y = x + 1 − v, that is,
y
=
=
=
Ce2x + 1
Ce2x − 1
Cxe2x − x + Ce2x − 1 − Ce2x − 1
Ce2x − 1
Cxe2x − x − 2
, for some nonzero constant C.
Ce2x − 1
x+1−
In summary, the general solution to y 0 = (x − y + 1)2 is:
either y = x, or y =
Cxe2x − x − 2
.
Ce2x − 1
Remark 1. The following are some general things to use in substitutions:
When you see
Try substituting
Examples of Differential equations
yy 0
v = y2
yy 0 + p(x)y 2 = q(x)
y2 y0
v = y3
y 2 y 0 + p(x)y 3 = q(x)
yα y0
v = y α+1
cos(y)y 0
v = sin(y)
cos(y)y 0 + p(x) sin(y) = q(x)
sin(y)y 0
v = cos(y)
sin(y)y 0 + p(x) sin(y) = q(x)
y 0 ey
v = ey
ey y 0 + p(x)ey = q(x)
for some constant α
y α y 0 + p(x)y α+1 = q(x)
BERNOULLI EQUATIONS
Let n be any fixed number, the Bernoulli equation is of the following form:
y 0 + p(x)y = q(x)y n .
a. If n = 0, then y 0 + p(x)y = q(x). Hence
r(x) = e
R
p(x) dx
,
y = e−
R
p(x) dx
Z
e
R
p(x) dx
q(x).
b. If n = 1, then y 0 + p(x)y = q(x)y, that is, y 0 = [q(x) − p(x)]y. Hence
Z
either y(x) ≡ 0, or ln |y| =
1
=
y
Z
[q(x) − p(x)] dx.
That is,
R
y = Ce
[q(x)−p(x)] dx
.
SUBSTITUTION
3
c. If n 6= 0, 1, then y −n y 0 + p(x)y 1−n = q(x). Let v = y 1−n , then
v 0 = (1 − n)y −n y 0 .
So
v0
+ p(x)v = q(x).
1−n
That is,
v 0 + (1 − n)p(x)v = (1 − n)q(x).
Then
r(x) = e
(1−n)
R
p(x) dx
,
v(x) = e
R
(n−1)
p(x) dx
Z
(1 − n)q(x)e(n−1)
R
p(x) dx
.
Since v = y 1−n , then
R
y = (1 − n)e(n−1) p(x)
dx
Z
q(x)e
(n−1)
R
p(x) dx
1
1−n
Example 2. Solve xy 0 + y(x + 1) + xy 5 = 0, y(1) = 1.
Rewrite the equation:
x+1
· y = −y 5 , y(1) = 1.
x
x+1
So it’s a Bernoulli equation with p(x) =
and q(x) = −1. Multiply y −5 on both sides of the equation, we get
x
y0 +
y −5 y 0 +
x + 1 −4
· y = −1.
x
Let v = y −4 , then v(1) = 1, and
v 0 = −4y −5 y 0 .
So
−
v0
x+1
+
· v = −1.
4
x
That is,
v0 −
4(x + 1)
· v = 4,
x
v(1) = 1.
Then
r(x) = e−
Rx
1
4(t+1)
t
dt
= e4−4(x+ln |x|) .
So
v=e
−4+4(x+ln |x|)
Z
x
4e
1
4−4(t+ln |t|)
dt + 1 .
4
MINGFENG ZHAO
Since v(1) = 1, that is, the domain interval of v should contain 1, so x > 0. Hence
Z x
−4+4(x+ln x)
4−4(t+ln t)
4e
v = e
dt + 1
1
=
x
Z
x4 e4x−4 4
t−4 e4−4t dt + 1 .
1
1
Since v = y −4 , that is, y = v − 4 . Hence, the solution to xy 0 + y(x + 1) + xy 5 = 0, y(1) = 1 is:
y=
e1−x
Rx
.
x 4 1 t−4 e4−4t dt + 1
Homogeneous equations
Let’s consider the following so called homogeneous equations:
y0 = F
Let v =
y
x
.
y
, that is, y = vx. So
x
y 0 = xv 0 + v = F (v).
That is,
v0 =
F (v) − v
.
x
Then
Z
either v(x) ≡ a for some constant a such that F (a) = a, or
Example 3. Solve x2 y 0 = y 2 + xy, y(1) = 1.
Rewrite the equations:
y0 =
Let v(x) =
y2
y
+ ,
2
x
x
y(1) = 1.
y
, then v(1) = 1, y = xv, and
x
y 0 = xv 0 + v = v 2 + v.
So
xv 0 = v 2 .
That is,
v0 =
v2
.
x
1
dv =
F (v) − v
Z
1
= ln |x| + C.
x
SUBSTITUTION
5
Since v(1) = 1, then v(x) 6≡ 0. So
Z
1
dv =
v2
Z
1
dx.
x
That is,
−
1
= ln |x| + C.
v
Since v(1) = 1, then −1 + 0 = C, that is, C = −1. Since v(1) = 1, that is, the domain interval of v should contain 1,
so x > 0. Hence
−
1
= ln x − 1.
v
Then
v=
Since v =
1
.
1 − ln x
y
, then the solution to is:
x
Example 4. Solve y 0 =
y = xv =
x
.
1 − ln x
Z
Z
x
.
y
Method I: Let f (x) = x and g(y) =
1
, then
y
y dy =
x dx.
That is,
y2
x2
=
+ C.
2
2
x
is:
y
Since C is arbitrary constant, then the general solution to y 0 =
x2 − y 2 = C .
Method II: Let v =
y
, then y = xv, that is,
x
y 0 = xv 0 + v =
1
.
v
So
v0 = −
v−
x
1
v
.
So either v = ±1, or
Z
1
v−
Z
1
v
dv =
1
dx = − ln |x| + C.
x
6
MINGFENG ZHAO
Z
For
1
v−
1
v
dv, we have
Z
1
v−
Z
1
v
dv
=
=
=
v
dv
v2 − 1
Z 1
1
1
+
dv
2
v+1 v−1
1
ln |(v + 1)(v − 1)| + C.
2
So we get
1
ln |(v + 1)(v − 1)| = − ln |x| + C.
2
That is,
(v + 1)(v − 1) = Cx−2 .
Since v =
y
, then
x
y
x
+1
y
x
− 1 = Cx−2 .
x
So the solution to y = is:
y
0
x2 − y 2 = C .
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca