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TECHNIQUES FOR SOLVING DIFFERENTIAL EQUATIONS
Separable Equations
dy dx
= g(x)f (y)
Step 1: Separate x’s and y’s on different sides.
f (y)
dy = g(x)dx
Step 2: Integrate both sides.
f (y)
dy =
g(x)dx + C
Step 3: Express y in terms of x where possible. If |y| = h(x), then y = ±h(x). If y = ±eC^ h(x), then y = A h(x) where A is any real number (including zero). Step 4: Check that constant solutions y = C where f (C) = 0 are not missed.
First Order Linear Equations y′^ + P (x)y = Q(x)
Step 1: Find integrating factor. I(x) = e
R P (x)dx
Step 2: Write differential equation as
I(x)y
= I(x)Q(x)
Step 3: Integrate both sides. I(x)y =
I(x)Q(x)dx + C
Step 4: Divide both sides by I(x) y =
I(x)
I(x)Q(x)dx + C
Remark: Be careful with the sign of P (x). For instance,
If y′^ + (^1) x y = 1, the integrating factor is I(x) = e
R (^1) /x dx = x. If y′^ − (^1) x y = 1, the integrating factor is I(x) = e
R (^) − 1 /x dx = (^1) x.
Second Order Linear Homogeneous Equations ay′′^ + by′^ + cy = 0
Step 1: Write down the auxiliary equation. ar^2 + br + c = 0
Step 2: Solve the auxiliary equation. r =
−b ±
b^2 − 4 ac 2 a
Step 3: (i) (ii) (iii)
Depending on the roots: b^2 − 4 ac > 0. Two real roots r 1 , r 2. b^2 − 4 ac = 0. One real root r = r 1 = r 2. b^2 − 4 ac < 0. Two complex roots α ± iβ.
y = c 1 er^1 x^ + c 2 er^2 x y = c 1 erx^ + c 2 x erx y = c 1 eαx^ cos βx + c 2 eαx^ sin βx
Second Order Linear Non-homogeneous Equations ay′′^ + by′^ + cy = G(x)
Method of Undetermined Coefficients
Step 1: Solve the complementary equation. ay c′′ + by′ c + cyc = 0
yc = c 1 y 1 (x) + c 2 y 2 (x)
Step 2: Write down a trial solution:
(i) G(x) = P (x) yp = Q(x)
(i) G(x) = P (x)ekx^ yp = Q(x) ekx
(i) G(x) = P (x)ekx^ cos mx or P (x)ekx^ sin mx yp = Q(x) ekx^ cos mx + R(x) ekx^ sin mx
Here, P (x), Q(x) and R(x) are polynomials of the same degree.
Multiply yp by x (or x^2 ) if one of the terms in the sum is y 1 (x) or y 2 (x).
Step 3: Substitute yp into the differential equation, group terms of the same form together, e.g. xnekx^ cos mx, xnekx^ sin mx and solve for the unknown coefficients.
ay p′′ + by′ p + cyp = G(x)
Step 4: Write down the general solution. y(x) = yc(x) + yp(x)
Second Order Linear Non-homogeneous Equations ay′′^ + by′^ + cy = G(x)
Method of Variation of Parameters
Step 1: Solve the complementary equation. ay c′′ + by′ c + cyc = 0
yc = c 1 y 1 + c 2 y 2
Step 2: The particular solution has the form: yp = u 1 y 1 + u 2 y 2
Write down the two conditions: u′ 1 y 1 + u′ 2 y 2 = 0 u′ 1 y 1 ′ + u′ 2 y′ 2 = G(x)/a
Solve the conditions for u′ 1 and u′ 2. u′ 1 =
G(x)y 2 a(y 1 ′y 2 − y′ 2 y 1 )
u′ 2 =
G(x)y 1 a(y′ 2 y 1 − y 1 ′y 2 )
Step 3: Integrate u′ 1 , u′ 2 to get u 1 , u 2. u 1 =
u′ 1 dx + c 1 u 2 =
u′ 2 dx + c 2
Step 4: Write down the general solution. y = (
u′ 1 dx + c 1 )y 1 + (
u′ 2 dx + c 2 )y 2
KEY
?^
Definitions
•^
Theorem
◦^
Remark
Extra
9.1 Modeling with Differential Equations
?^
differential equation, order.solution, general solution. ?^
equilibrium solution: a constant solution
y
C
set
y
′′^
y
′^ = 0
, y
C
in diff eqn and solve for
C
?^
initial condition, initial value problem.
9.2 Direction Fields and Euler’s Method
?^
direction field, solution curve. ?^
autonomous differential equation
y
′^
f
(y
if y
g
(x
) is a solution, so is
y
g
(x
C
e.g. natural growth, logistic model ◦^
Graphical method:1. draw direction field.2. draw solution curve. ◦^
Numerical method: Euler’s method, step size
h
Solving
y
′^ =
F
(x, y
), y
(x
y
- Set
x
n^
x
0
nh
for
n
- Recursively,
y
n+
y
n^
hF
(x
, yn
) forn
n
9.3 Separable Equations
?^
separable equations, orthogonal trajectories,mixing problems (see formula sheet)
9.4 Population Models
◦^
know how to derive solutions of natural growth/logistic modelusing separation of variables and partial fractions. ?^
law of natural growth (see formula sheet)compare with exponential decay
P
′^
kP
where
k
is negative,
P
(t
P
e 0 − kt
?^
logistic differential equation (see formula sheet)case 1: 0
< P
0
< K
.^
P
increases and approaches
K
case 2:
P
0
> K
.^
P
decreases and approaches
K
know how to see this from the differential equation. ⊗
natural growth with harvesting
P
′^
kP
c, P
P
0
P
(t
)^
c k
P
0
c k )e
kt
trick: subs
y
P
c k
to get natural growth model.
logistic model with harvesting (see quiz 11)
P
′^
kP
K P
c, P
P
0
two equilibrium solutions
P
, P 1
2
K 2
(^4) ckK
case 1:
P
0
< P
P
approaches
case 2:
P
1
< P
0
< P
P
increases and approaches
P
case 3:
P
2
< P
P
decreases and approaches
P
trick: subs
y
P
P
1
to get logistic model.
Seasonal growth
P
′^
kP
cos(
rt
φ
P
Ce
(k/r
) sin(
rt −φ
)
Seasonal growth with harvesting
P
′^
kP
cos(
rt
φ
)^
c
P
′^
kP
P K
m P
m
extinction level.
9.5 First Order Linear Equations
?^
first order linear equation (see formula sheet) ◦^
to get unique solution from general solution:initial value problem
y
y
0
9.6 Predator-Prey Systems
?^
predator prey equations (see formula sheet) ?^
phase plane, phase trajectory, phase portrait ?^
know how to derive, draw and comparephase trajectories and population graphs ⊗
dWdR
rW
bRW
kR
aRW
R
rW
k
bR e
aWe
C
(see S9.6Q7)
11.1 Second Order Linear Equations
?^
second order linear equations^ P
(x
)y
′′^
Q
(x
)y
′^ +
R
(x
) = 0 homogeneous
P
(x
)y
′′^
Q
(x
)y
′^ +
R
(x
G
(x
) nonhomogeneous
?^
linear combination, linearly independent solutions
-^
if differention equation is linear and homoegeneous,then linear combinations of solutions are solutions. ◦^
P
(x
)y
′′^
Q
(x
)y
′^ +
R
(x
G
(x
to get unique solution from general solution:1. initial condition, initial value problem
y
y
, y 0
y
′ 0
always has unique solution near
x
if
P, Q, R, G
continuous and
P
- boundary condition, boundary value problem
y
y
, y 0
y
1
may not have a solution
?^
second order linear homogeneous equation,auxiliary/characteristic equation (see formula sheet)
11.2 Nonhomogeneous Linear Equations
?^
complimentary equation
ay
′′^
by
′^ +
cy
complimentary solution
y
(c x
particular solution
y
(p x
general solution
y
(x
y
(p x
y
(c x)
?^
method of undetermined coefficients (see formula sheet) ?^
variation of parameters (see formula sheet)
11.3 Applications of Second Order Differential Equations
?^
spring systems (see formula sheet)simple harmonic motiondamped vibrations
overdamping (returns to rest slowly)critical damping (returns to rest fastest)underdamping (oscillates before coming to rest)know how to draw graphs of above casesknow how to check if graph cuts
x
-axis
forced vibrations item[
?] electric circuits (see formula sheet)
?^
steady state solutionis behavior of solution as
t
see S17.3 example 3 note 1 ?^
a function
x
(t
) has period
T
if
x
(t
T
x
(t
) for all
t
11.4 Series Solutions
◦^
finds solutions near
x
= 0 because we are writing
the solution as a power series with center
x
◦^
step 1: write
y
∞ n=
c
xn n.
step 2: differentiate to get
y
′,^
′′y as power series.
step 3: substitute into differential equation. collect terms.step 4: equate coefficients to get
recursion relations
step 5: solve recursion relations for small
n
to find patterns.
step 6: write down general solution.
it will be in terms of certain coefficientse.g.
c
, c 0
, which act as arbitrary constants. 1
step 7: (optional) recognize terms in general solution
as power series of well-known functions.
