Boyce\'s Elementary Differential Equations and Boundary Value Problems
Summary study book Boyce's Elementary Differential Equations and Boundary Value Problems of William E. Boyce, Richard C. DiPrima (Chapter 2, 3, 5, 7, 9, 10) - ISBN: 9781119382874
Very clear summary with many examples and step-by-step explanation of the fabric
Seller
Follow
rhjatol
Reviews received
Content preview
Summary Differential Equations
by Ruben Tol
Ch. 2 First-Order Differential Equations
Standard form
dy
= f (t, y)
dx
§2.1 Linear Differential Equations; Method of Integrating
Factors
Standard form
dy
+ p(t)y = g(t), y(t0 ) = y0
dt
How to solve
R
p(t) dt
(1) Multiply both sides of the equation by the integrating factor µ(t) = e :
dy
µ(t) + µ(t)p(t)y = µ(t)g(t).
dt
(2) Rewrite the left-hand side of the equation using the reverse chain rule:
d
(µ(t)y) = µ(t)g(t).
dt
(3) Integrate both sides of the equation:
Z
µ(t)y = µ(t)g(t) dt + c.
(4) Divide both sides of the equation by the integrating factor µ(t):
Z
1
y= µ(t)g(t) dt + c .
µ(t)
(5) Solve y for t0 to obtain c.
Example
ty 0 + 2y = 4t2 , y(1) = 2
2
y 0 + y = 4t
t
2 R 2
p(t) = =⇒ µ(t) = e t dt = e2 ln |t| = t2
t
t2 y 0 + 2ty = 4t3 (1)
1
, d 2
t y = 4t3 (2)
dtZ
t2 y = 4t3 dt = t4 + c (3)
c
y = t2 + (4)
t2
c
y(1) = 2 =⇒ 2 = 12 + =⇒ c = 1 (5)
12
1
y = t2 +
t2
§2.2 Separable Differential Equations
Standard form
dy
N 0 (x, y) + M 0 (x, y) = 0, y(x0 ) = y0
dx
How to solve
(1) Try to rewrite the equation by rewriting M (x, y) to a function of only x and
N (x, y) to a function of only y, by pulling all functions of x to one side, and all
functions of y to the other side:
dy
N 0 (y)
= M 0 (x).
dx
(2) Write the equation in differential form:
N 0 (y) dy = M 0 (x) dx.
(3) Integrate both sides of the equation:
N (y) = M (x) + c.
(4) Solve y for t0 to obtain c.
Example
dy 3x2 + 4x + 2
= , y(0) = −1
dx 2(y − 1)
dy
2(y − 1) = 3x2 + 4x + 2 (1)
dx
2(y − 1) dy = (3x2 + 4x + 2) dx (2)
Z Z
2(y − 1) dy = 3x2 + 4x + 2 dx =⇒ y 2 − 2y = x3 + 2x2 + 2x + c (3)
(y − 1)2 = x3 + 2x2 + 2x + 1 + c
√
y = 1 ± x3 + 2x2 + 2x + 1 + c
√
y(0) = 1 =⇒ 1 = 1 ± 1 + c (4)
√ √
1 + c > 0 =⇒ −1 = 1 − 1 + c =⇒ c = 3
√
y = 1 − x3 + 2x2 + 2x + 4
2
,§2.4 Differences Between Linear and Nonlinear Differential
Equations
df
Let f (t, y) and dy be continuous in some rectangle α < t < β, γ < y < δ containing
the initial value (t0 , y0 ). Then, in some interval t0 − h < t < t0 + h ∈ [α, β] there is
a unique solution y = φ(t) to the initial value problem.
Example
Let’s take the example equation from §2.1,
ty 0 + 2y = 4t2 .
This equation has the solution
1
y = t2 + .
t2
This solution is only continuous on the interval −∞ < t < 0, 0 < t < ∞ (t 6= 0).
Now, if we take the initial condition to be y(1) = 2, the unique solution φ(t) lays
in the interval 0 < t < ∞, for t0 = 1 ∈ [0, ∞]. If we take the initial condition
to be y(−1) = 2, the unique solution φ(t) lays in the interval −∞ < t < 0, for
t0 = −1 ∈ [−∞, 0].
§2.5 Autonomous Differential Equations and Population Dy-
namics
Standard form
dy
= ry
dx
Here, r is called the rate of growth if r > 0, or the rate of decline if r < 0.
Solving for the initial condition y(0) = y0 yields
y = y0 ert .
A population growth can be modelled with this equation, where y0 equals the amount
of a species at t0 and r > 0, allowing this population to grow exponentially.
dy
To show this growth, we plot dx , y and indicate the direction of the growth using
arrows on the y-axis. This line is called the phase line. For this equation, the phase
line is
0 −→ ∞.
To account for the fact that the growth rate actually depends on the population, we
will rewrite this equation to
dy y
=r 1− y.
dt K
Here, r is the intrinsic growth rate, that is, the growth rate in the absence of any
limiting factor, and K is the saturation level, that is, the environmental carrying
capacity for the given species, both positive constants.
3
, dy
Solving this equation for dx = 0, when the population growth is stable, yields the
constant solutions y = φ1 (t) = 0 and y = φ2 (t) = K. These solutions are called the
equilibrium solutions, with 0 and K being called the critical points of the equation:
the function will eventually converge to y = φ2 (t) = K. φ1 (t) = 0 is called the
unstable equilibrium solution, meaning the population died out. y = φ2 (t) = K is
called the asymptotically stable solution, meaning the population will get closer and
closer to y = K:
y0 K
lim y(t) = = K.
x→∞ y0
The phase line for this equation contains the critical point K:
0 −→ K ←− ∞.
Now, let’s consider the following equation:
dy y
= −r 1 − y.
dt T
Here, T is the threshold level, also a positive constant.
dy
Solving this equation for dx = 0, when the population growth is stable, yields the
constant solutions y = φ1 (t) = 0 and y = φ2 (t) = T . Now, if 0 < y < T , then
dy
dt
< 0, implying the population dies out over time and approaches the solution
y = φ1 (t) = 0. On the other hand, if y > T , then dy
dt
< 0, implying the population
keeps growing exponentially, with y = φ2 (t) = T being the unstable equilibrium
solution.
This equation has the following phase line:
0 ←− T −→ ∞.
To get rid of any unbounded growth, the following equation is used:
dy y y
= −r 1 − 1− y.
dt T K
where r is a positive constant and 0 < T < K.
Again, solving for dy
dt
now yields three critical points: y = φ1 (t) = 0, y = φ2 (t) = T
and y = φ3 (t) = K. Now, if T < y < K, then dy dt
> 0, allowing growth. But, if y < T
dy
or y > K, dx < 0, decreasing the population size in those intervals. Consequently,
the equilibrium solutions y = φ1 (t) = 0 and y = φ3 (t) = K are asymptotically
stable, and the solution y = φ2 (t) = T is unstable. This equation creates the
following population growth:
If y > K, the population is too big to be sustained, and declines down towards the
asymptotically stable solutions y = φ3 (t) = K.
If y < T , the population is too small to be sustained, and declines down towards
the asymptotically stable solutions y = φ1 (t) = 0.
4
, If T < y < K, the population is big enough to grow, inclining towards the asymp-
totically stable solution y = φ3 (t) = K and away from the unstable equilibrium
solution y = φ2 (t) = T .
This yields the following phase line:
0 ←− T −→ K ←− ∞.
§2.6 Exact Differential Equations and Integrating Factors
Standard form
∂N (x, y) ∂M (x, y)
N (x, y)y 0 + M (x, y) = 0, where =
∂x ∂y
How to solve
(1) Define a function ψ(x, y) with the following property:
∂ψ ∂ψ
N (x, y) = , M (x, y) = .
∂y ∂x
(2) Rewrite the equation to
∂ψ ∂ψ dy
+ = 0.
∂x ∂y dx
(3) Rewrite the left-hand side of the equation using the chain rule:
d
ψ(x, y) = 0.
dx
(4) Integrate both sides of the equation:
ψ(x, y) = c.
Solutions of the equation are defined implicitly by ψ(x, y) = c.
Example
2x + y 2 + 2xyy 0 = 0
∂ψ ∂ψ
2x + y 2 = , 2xy = =⇒ ψ(x, y) = x2 + xy 2 (1)
∂x ∂y
∂ψ ∂ψ dy d d 2
+ = 0 =⇒ ψ(x, y) = 0 =⇒ (x + xy 2 ) = 0 (2,3)
∂x ∂y dx dx dx
ψ(x, y) = x2 + xy 2 = c (4)
5
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller rhjatol. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $6.96. You're not tied to anything after your purchase.
