MTH306 Notes – May 29
Section 1.6 Bernoulli equations
• Differential equations of form
dy
dx
= F (ax + b y + c).
– Use substitution v = ax + b y + c.
• Homogeneous equations
dy
dx
y
= F ( x ).
y
– Use substitution v = x .
• Bernoulli equations
dy
dx
+ P(x) y = Q(x) y n .
– n = 0: Standard linear first order equation. (Section 1.5)
– n = 1:
dy
dx
= (Q(x) − P(x)) y. Separable equation. (Section 1.4)
– all the other n: do the substitution v = y 1−n .
Steps:
(1) Substitution: v = y 1−n . Then
dv
= (1 − n) y −n
dx
(2) Multiply the original equation by (1 − n) y −n . We have
dv
dx
dy
dx
+ (1 − n)P(x) v = (1 − n)Q(x)
|
|
{z
}
{z
}
new P(x)
(3) Solve (2) for v(x) and get the implicit solution for y.
new Q(x)
Example. §2.6 # 7 Solve the equation x y 2 y 0 = x 3 + y 3 .
Solution: (1) Rearrange the equation:
dy
dx
=
x3
x y2
+
y3
x y2
1
+ (− ) y = x 2 y −2 ,
dx
x
dy
(2) Substitution: v = y 1−n = y 3 . Then
= x 2 y −2 + 1x y. That is
–Bernoulli equation., n = −2
dy
= 3 y2 dx .
dv
dx
(3) Multiply the equation by 3 y 2 , we get
3 y2
That is
1
+ 3 y 2 · (− ) · y = 3 y 2 · x 2 y −2
dx
x
dy
3
+ (− )v = 3x 2
dx
x
(3) Solve this equation and get the solution.
dv
–Standard linear 1st order equation
y 3 = v = x 3 (3 ln |x| + C).
Equation
dv
dx
+ (− 3x )v = 3x 2 . Then P(x) = − 3x , Q(x) = 3x 2 .
(1) Integrating factor:
ρ(x) = e
R
− 3x d x
= e−3 ln x = x −3 =
1
x3
(2) Multiply both sides by the integrating factor, we have
Z
Z
1
1
3
2
v(x) =
· 3x d x =
d x = 3 ln |x| + C
3
3
x
x
x
(3) Solve for v:
y 3 = v = x 3 (3 ln |x| + C)
Section 2.2 Equilibrium solutions and stability.
• Autonomous equation:
dx
dt
= f (x). (x is a function of t.)
• Equilibrium solutions: the solution of f (x) = 0. (Critical points)
• Stability: Suppose x = c is a critical point.
- Stable: Solution curves approach x = c as t → +∞.
- Unstable: Solution curves diverge away from x = c as t → +∞.
• Steps to find the equilibrium solutions and stability:
(1) Find critical points of autonomous 1st order ODE;
(2) Classify critical points by analyzing sign of f (x) around critical points.
(Pick points around the critical points.)
(3) Draw the phase diagram.
Example. §2.2 # 1
dx
dt
= x − 4.
Solution: f (x) = x − 4. This is an autonomous equation.
(1) Critical points: f (x) = x − 4 = 0 ⇒ x = 4.
(2) Stability at x = 4:
If x > 4 and x is close to 4, f (x) = x − 4 > 0. Then x(t) %; (We can pick x = 5.)
If x < 4 and x is close to 4, f (x) = x − 4 < 0. Then x(t) &. (We can pick x = 3.)
Then x = 4 is unstable.
(3) Phase diagram:
x = 4 is unstable:
Example. §2.2 # 6
dx
dt
= 9 − x 2.
Solution: f (x) = 9 − x 2 . This is an autonomous equation.
(1) Critical points: f (x) = 9 − x 2 = 0 ⇒ x = 3, −3.
(2) Stability at x = 3:
If x > 3 and x is close to 3, f (x) = 9 − x 2 < 0. Then x(t) &; (We can pick x = 4.)
If x < 3 and x is close to 3, f (x) = 9 − x 2 > 0. Then x(t) %. (We can pick x = 2.)
Then x = 3 is stable.
Stability at x = −3:
If x > −3 and x is close to −3, f (x) = 9 − x 2 > 0. Then x(t) %; (We can pick x = −2.)
If x < −3 and x is close to −3, f (x) = 9 − x 2 < 0. Then x(t) &. (We can pick x = −4.)
Then x = −3 is unstable.
(3) Phase diagram:
x = 3 is stable; x = −3 is unstable.
a