Lecture 28: Solving First-Order DEs by
Substitution Methods:
Homogeneous DEs, Bernoulli DEs,
and Review of First-Order Linear DEs
Winfried Just
Department of Mathematics, Ohio University
November 18, 2015
Winfried Just
MATH3400, Lecture 28: Substitution Methods
A homogeneous first-order DE
Example 1: Consider the DE
dy
dx
y
= e x + yx .
This is not a separable DE, but its right-hand side depends only on
the ratio yx .
y
x
A DE of the form dy
dx = F ( x ) = G ( y ) is called a first-order
homogeneous DE.
To solve such DEs, we can make the substitution u = yx .
Then y = xu, and the product rule gives
dy
dx
= u + x du
dx .
u
In the new variable, the DE becomes u + x du
dx = e + u, or
u
x du
dx = e .
The original homogeneous DE turns into the separable DE
du
1 u
dx = x e .
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
A homogeneous DE: Example 1 continued
Consider the DE
dy
dx
y
= e x + yx .
After the substitution u =
the separable DE
du
dx
y
x
=
this DE turns into
eu
x .
Step 1: Separate variables:
R
Step 2: Form integrals:
Step 3: Integrate both sides:
Step 4: Solve for u:
dx
du
eu = x .
R
e −u du = x1 dx.
−e −u = ln |x|
+ C.
u = − ln(C − ln |x|).
Step 5: Express the solution in terms of the original variable y :
y (x) = xu(x) == −x ln(C − ln |x|).
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Another homogeneous first-order DE
Example 2: Consider the DE (x 2 + y 2 ) dy
dx = 2xy +
y3
x .
1 1
−1 y 3 .
This DE can be written as (x 2 y 0 + x 0 y 2 ) dy
dx = 2x y + x
The exponents add up to the same number for each term, which
indicates that this DE is homogeneous.
Divide both sides by (x 2 + y 2 ) and then simplify:
dy
dx
=
2xy
x 2 +y 2
=
x 2 +y 2
xy
+
+
y3
x(x 2 +y 2 )
y (x 2 +y 2 )
x(x 2 +y 2 )
=
=
y
x
xy
x 2 +y 2
2
1+ y 2
x
+
y
x
After the substitution u = yx ,
we get the separable DE
or
du
dx
=
+
x 2y
x(x 2 +y 2 )
=
y
x
1+( yx )
y = xu,
u + x du
dx =
2
+
y3
x(x 2 +y 2 )
+ yx .
dy
dx
u
1+u 2
= u + x du
dx
+ u,
u
.
x(1+u 2 )
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Another homogeneous DE: Example 2 continued
(x 2 + y 2 ) dy
dx = 2xy +
Example 2: Consider the DE
After the substitution u =
the separable DE
du
dx
y
x
=
y3
x .
this DE turns into
u
.
x(1+u 2 )
Step 1: Separate variables:
Step 2: Form integrals:
( u1 + u)du = dx
x .
R 1
R 1
u + u du =
x dx.
Step 3: Integrate both sides:
ln |u| +
u2
2
= ln |x| + C .
Step 4: Solving explicitly for u is difficult here. But the previous
step already gives us a family of implicit solutions.
Step 5: Express these implicit solutions in terms of the original
variable y :
y2
ln yx + 2x
2 = ln |x| + C .
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
What is an implicit solution, really?
Definition
An equation G (x, y ) = 0 is said to be an implicit solution of an
ODE on an interval I , provided that there exists at least one
function y (x) that satisfies the relation as well as the differential
equation on I .
For example, the equation G (x, y ) = x 2 + y 2 − 25 = 0
implicitly defines two differentiable functions
√
√
y1 (x) = 25 − x 2 and y2 (x) = − 25 − x 2
on the interval I = (−5, 5).
Homework 37: Verify that both of these functions are solutions
of the DE
dy
dx
=
−x
y
on I = (−5, 5).
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Example 3: A Bernoulli DE
Example 3: The DE
dy
dx
+
y
x
= xy 2
is neither separable, homogeneous, nor linear.
It is an example of a Bernoulli equation, which are of the form
dy
dx
+ P(x)y = Q(x)y n for some n 6= 0, 1.
To solve Bernoulli DEs, we can make the substitution u = y 1−n .
1
Then y = u 1−n , and by the Chain Rule,
dy
dx
=
n
1
1−n du .
1−n u
dx
In our example, n = 2, u = y −1 , y = u −1 , and
dy
dx
= −u −2 du
dx .
Substituting these expressions in the DE gives
−u −2 du
dx +
11
xu
= xu −2 .
After dividing both sides by −u −2 we get the linear DE
du
dx
− ux = −x. The substitution u = y 1−n always reduces a
Bernoulli DE to a linear DE.
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Example 3 completed: Solving a linear DE
The substitution u = y −1 reduces the Bernoulli DE
dy
dx
+
y
x
= xy 2 to the linear DE
du
dx
−
u
x
= −x.
du
dx + P(x)u
R
e P(x) dx .
Recall the steps for solving a linear DE
= Q(x):
Step 1: Find the integrating factor µ =
In our example, P(x) = − x1 and µ = e − ln |x| = x1 .
Step 2: Multiply both sides of the DE by µ.
The left-hand side becomes (µu)0 .
0
In our example, x1 u 0 − x12 u = ux = x1 (−x) = −1.
Step 3: Integrate the right-hand side.
R
In our example, ux = −1 dx = −x + C .
Step 4: Divide both sides by µ. Here we get: u(x) = −x 2 + Cx.
Step 5: Express the solution of the Bernoulli DE in terms of
the original variable y :
y (x) =
Ohio University – Since 1804
1
u(x)
=
1
.
−x 2 +Cx
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Another Bernoulli DE: Example 4
dy
dx
Example 4: The DE
+ 23 y tan x =
sec
√x
y
is another example of a Bernoulli equation, which are of the form
dy
dx
+ P(x)y = Q(x)y n for some n 6= 0, 1.
To solve Bernoulli DEs, we can make the substitution u = y 1−n .
1
Then y = u 1−n , and by the Chain Rule,
dy
dx
=
n
1
1−n du .
1−n u
dx
In our example, n = − 21 , u = y 3/2 , y = u 2/3 ,
dy
dx
= 32 u −1/3 du
dx .
Substituting these expressions in the DE gives
2 −1/3 du
3u
dx
+ 32 u 2/3 tan x =
sec x
.
u 1/3
After dividing both sides by 23 u −1/3 we get the linear DE
du
dx
+ u tan x = 32 sec x. The substitution u = y 1−n always
reduces a Bernoulli DE to a linear DE.
Ohio University – Since 1804
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods
Example 4 completed: Solving a linear DE
The substitution u = y 3/2 reduces the Bernoulli DE
dy
dx
+ 32 y tan x =
sec
√x
y
du
dx
to the linear DE
du
dx + P(x)u
R
e P(x) dx .
Recall the steps for solving a linear DE
Step 1: Find the integrating factor µ =
In our example, P(x) = tan x and µ = e
R
+ u tan x =
tan x dx
3
2
sec x.
= Q(x):
= e − ln | cos x| =
1
cos x .
Step 2: Multiply both sides of the DE by µ.
The left-hand side becomes (µu)0 , so that in our example
0
1
sin x
u
0
= cos1 x 32 sec x = 32 sec2 x.
cos x u + cos2 x u = cos x
Step 3: Integrate the right-hand side.
R
In our example, cosu x = 32 sec2 x dx =
3
2
tan x + C .
Step 4: Divide both sides by µ. We get: u(x) =
Step 5: Express in terms of y : y (x) =
Ohio University – Since 1804
( 23
3
2
sin x + C cos x.
sin x + C cos x)2/3 .
Department of Mathematics
Winfried Just
MATH3400, Lecture 28: Substitution Methods