AMS361 Recitation 03/04 Notes 4
1
The Exact DEs and Almost Exact DEs
General Formula:
M (x, y)dx + N (x, y)dy = 0
1.1
Exact
If My = Nx , then the DE is called exact. Furthermore, the general solution is given by
F (x, y) = C. Do the total differentiation, we have
Fx dx + Fy dy = 0.
1.1.1
Steps for Solving
• SETP1: Compare this equation with the general formula. If we want two match, then
Fx = M (x, y),
(1)
Fy = N (x, y).
(2)
• SETP2: Look at the equation (1), it can be solved by direct integration.
Z
Z
Fx dx = M (x, y)dx
Z
F (x, y) = M (x, y)dx + g(y)
(3)
Note: Or, you can look at the equation (2), it can also be solved by direct
integration.
Z
Z
Fy dy = N (x, y)dy
(4)
Z
F (x, y) = N (x, y)dx + g(x)
• STEP3: Then, plug (3) into (2) to obtain g(y).
Note: Or, plug (4) into (1) to obtain g(x).
1
• SETP4: Finally, the solution is
Z
M (x, y)dx + g(y) = C.
Note: Or the solution is
Z
N (x, y)dy + g(x) = C.
1.1.2
Example 1
Question: Solving the following DE using the exact DE method
[1 + ln (xy)]dx +
x
dy = 0
y
Solution: First of all, we need to identify what is M (x, y) and N (x, y).
M (x, y) = 1 + ln (xy)
x
N (x, y) =
y
Then calculate My and Nx .
1
= Nx
y
My =
This tells that the DE is exact. Then we can follow the above steps.
• STEP1:
Fx = M (x, y) = 1 + ln (xy),
x
Fy = N (x, y) =
y
• STEP2:
Z
F (x, y) =
M (x, y)dx + g(y)
Z
=
1 + ln (xy)dx + g(y)
= x ln (xy) + g(y)
• STEP3:
x
x
+ g 0 (y) =
y
y
0
⇒ g (y) ⇒ g(y) = C
Fy =
• STEP4: The solution is
x ln (xy) = C.
2
1.2
Not Exact
Then, can we still the DE if it is not exact? For two special cases, it can still be solved by
integrating factor.
• 1. If
My −Nx
N
= f (x), then the integrating factor is ρ(x) = e
• 2. If
Nx −My
M
= g(y), then the integrating factor is ρ(y) = e
R
f (x)dx ;
R
g(y)dy .
Then, the original DE, which is not exact, can be converted to an exact DE by multiplying
the integrating factor ρ.
1.2.1
Example 2
Question: Solving the following DE using the exact DE method
(2x − y 2 )dx + xydy = 0
Solution: First, let us check if the DE is exact. We have
M (x, y) = 2x − y 2 , ⇒ My = −2y;
N (x, y) = xy, ⇒ Nx = y.
Obviously, My 6= Nx , the DE is not exact. However, we should notice the following.
My − Nx = −3y
My − Nx
3
=−
⇒
N
x
R
This is a single-variable function of x. Let ρ(x) = e
the original DE.
− x3 dx
= x−3 and then multiply it to
2x − y 2
y
dx + 2 dy = 0
3
x
x
We know the above DE is exact. Then we can follow the steps for exact DEs.
Z
Z
y
y2
F (x, y) = N (x, y)dy =
dy
=
+ g(x)
x2
2x2
2x − y 2
y2
⇒ M (x, y) = Fx = − 3 + g 0 (x) =
x
x3
2
⇒ g 0 (x) = 2
x
2
⇒ g(x) = − + C
x
y2
2
⇒ F (x, y) = 2 −
2x
x
Therefore, the solution is
y2
2
− = C.
2
2x
x
3
2
2.1
General Formulas in Chapter 1
Direct Integral
y 0 = f (x)
Z
⇒ y(x) = f (x)dx + C
2.2
Separation of Variables
f (x)
y0 =
g(y)
Z
Z
⇒
g(y)dy = f (x)dx
2.3
Integrating Factors
y 0 + P (x)y = Q(x)
d
⇒
(ρ(x)y) = Q(x)ρ(x)
dx
Z
⇒ ρ(x)y =
Q(x)ρ(x)dx + C
where the integrating factor is defined as
R
ρ(x) = e
2.4
P (x)dx
.
Polynomial Substitution
y 0 = F (ax + by + c)
Introduce a new variable v = ax + by + c.
2.5
y
x
Substitution
y
y0 = F ( )
x
Introduce a new variable v = xy .
2.6
Bernoulli DEs
y 0 + P (x)y = Q(x)y n
Introduce a new variable v = y 1−n .
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