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Category > Math Posted 21 Apr 2017 My Price 12.00

MATH 2255: Differential Equations Prof

Please help, Urgent!!!!!!!!

Please help me to do these tow docs' HW.

 

 

MATH 2255: Differential Equations
Non-Homogeneous Linear Equations: Variation of Parameters Prof. ∀ll
June 29, 2016 Consider the linear non-homogeneous second order Example: Use the method of variation of parameters to
differential equation
find the general solution to y ′′ − 2y ′ − 3y = cos(2t).
(1) y ′′ + p(x) · y ′ + q(x) · y = g(x). We assume that p, q, g are continuous on some interval I
so that we can guarantee the existence of solutions. The
general solution to this equation is of the form y(x) =
yh (x) + yp (x) where yh (x) is the general solution to the
associated homogeneous equation and yp (x) is a particular solution. If p(x) and q(x) are constant functions, then
we’ve seen how the method of undetermined coefficients
can be effective in obtaining yp (x). On the other hand,
there are some obvious limitations to this method (e.g.,
how well can you guess at the form of the particular solution, etc).
When the method of undetermined coefficients seems
too unwieldily to apply, we can turn to another method.
Much like the method of reduction of order, we start by
assuming that the particular solution yp is some variable
combination of a fundamental set of solutions to the associated homogeneous equation, say y1 , y2 (which must
be known ahead of time):
yp (x) = c1 (x) · y1 (x) + c2 (x) · y2 (x).
After plugging this expression into 1 and simplifying, we
arrive at a pair of equations
0 = c′1 (x)y1 (x) + c′2 (x)y2 (x)
g(x) = c′1 (x)y1′ (x) + c′2 (x)y2′ (x)
In matrix form, we have
[ [ [ ′
0
y1 (x) y2 (x)
c1 (x)
= ′
· ′
.
g(x)
y1 (x) y2′ (x)
c2 (x)
Using the inverse of the matrix above we obtain
[ ′
c1 (x)
=
c′2 (x)
Upon integrating, we come to the following important
idea:
Variation of Parameters
Let y1 and y2 be a fundamental set of solutions to
the associated homogeneous equation for 1. Then
the particular solution yp (x) to 1 is given by
∫ x
∫ x
y2 (t)g(t)
y1 (t)g(t)
−y1 (x)
dt+y2 (x)
dt,
W (t)
W (t)
x0
x0
where W is the Wronskian of y1 , y2 and x0 is a
conveniently chosen constant (typically, x0 is chosen to match the initial condition if one is present). 2 HW: Due Wednesday, 6-Jul
1. Find the general solution to y ′′ + 4y = 3 csc(2x) on the interval 0 < x < π/2. 2. Consider the differential equation x2 y ′′ − x(x + 2)y ′ + (x + 2)y = 2x3 with x > 0. Use the fact that y1 (x) = x
and y2 (x) = xex form a fundamental set of solutions to the associated homogeneous equation to find a particular
solution to the ODE under consideration.

 

 

MATH 2255: Differential Equations Prof. ∀ll
June 29, 2016 Mechanical Vibrations
General setup. The ODE dx
d2 x
+ 2λ
+ ω2 x = f(t)
2
dt
dt
models the position x(t) of a mass attached to a spring under the influence of a variable force equivalent
to m · f(t) Newtons where
• m is the mass in kilograms
• k > 0 is a constant dependent on the spring in Newtons per meter
• β > 0 is a constant of proportionality dependent on the medium
• ω2 = k/m
• 2λ = β/m. (1) Free motion. This means that f(t) = 0 for all t. The general solution to this homogeneous equation
is,
√
as usual, determined by the roots of r2 + 2λr + ω2 = 0. These roots are precisely r = −λ ± λ2 − ω2 .
The nature of the solutions depends on how λ relates to ω. There are four cases to consider and they are
described below. The fundamental set of solutions x1 and x2 is given in each case. No damping:
λ=0 Under-damped:
0 < λ2 < ω2 Critically-damped:
λ2 = ω2 Over-damped:
λ2 > ω2 b b
b
b x1 = cos ωt
x2 = sin ωt √
x1 = e−λt cos √ ω2 − λ2 t
x2 = e−λt sin ω2 − λ2 t x1 = e−λt
x2 = te−λt √ 2 2 x1 = e(−λ−√λ −ω )t
2
2
x2 = e(−λ+ λ −ω )t Forced motion. Now assume f(t) 6= 0. Then by the method of undetermined coefficients, we determine a
particular solution xp to eq. (1). The complimentary solution xc is described in the previous section, so
the general solution is then x = xp + xc . If there’s an additive portion of x that goes to zero as t → ∞,
then that portion is called the transient solution, and whatever remains is called the steady-state solution.
Example: Suppose we have a 1 kg mass at the end of a spring that requires 2 N of force to stretch 1 meter
beyond its natural length. At t = 0 the spring has initial position x = 1 and initial velocity x ′ = 0. The
spring is under the influence of a variable external force of cos(2t) N and is submersed in a medium that
offers a resistive force numerically equivalent to twice the velocity of the mass. We want to determine the
position of the mass at any future time.
Since 2 = k · 1, it follows that k = ω2 = 2. Also, 2λ = 2/1 so λ = 1. Since λ2 < ω2 , the free-motion of
the spring is under-damped. So the solution to the homogeneous equation is
x = c1 e−t cos t + c2 e−t sin t. 2 Given the form of the function describing the forced motion, we guess that xp = A cos(2t) + B sin(2t) is a
particular solution to the non-homogeneous equation. We solve for the coefficients A and B through the
equality xp′′ + 2xp′ + 2xp = cos(2t) and find that A = −1/10 and B = 1/5. So the general solution is
− cos(2t) sin(2t)
+
+ c1 e−t cos(t) + c2 e−t sin(t).
10
5
Using the initial conditions x(0) = 1 and x ′ (0) = 0, we solve for c1 and c2 obtaining
x= x(t) = 7
− cos(2t) sin(2t) 11 −t
+
+ e cos(t) + e−t sin(t).
10
5
10
10
Solution bb HW: Due Wednesday, 6-Jul
1. Suppose a 1 kg mass is attached to a spring that requires 4 N of force to stretch 1 m beyond its natural
length, and assume the system is undamped. Now suppose that, from rest at the equilibrium position,
an external force acts on the spring-mass system. What’s the furthest the mass will be displaced from
equilibrium if that force equals sin(ηt) if η equals 1.7, 1.8, and 1.9? What if η = 2? 3 2. In mechanical systems, forced frequencies can sometime combine with the natural resonance of the
system to produce destructive oscillation. In practice, mechanical systems are almost always damped
to some degree. This keeps the amplitude of the oscillation from growing out of control regardless of
the frequency of forced oscillation. That being said, the forced frequency can still be dialed into potentially dangerous frequencies that cause large (albeit bounded) oscillations. An over-damped system
with damping coefficient 2λ and natural resonance ω subjected to a forced oscillation of frequency η
and amplitude F is modelled by
(2) x ′′ + 2λx ′ + ω2 x = F sin(ηt) λ < ω. The object of this problem is to determine which values of η produce the largest (sustained) oscillations.
(a) Let xp (t) be a particular solution to the above ODE, and let xc (t) be the general solution to the
homogeneous equation x ′′ + 2λx ′ + ω2 x = 0. Briefly explain why the general solution to 2 is
x(t) = xp (t) + xc (t). (b) Suppose the solution x to 2 can be written as x(t) = xs (t)+x0 (t) where x0 (t) → 0 as t → ∞. We call
x0 the transient solution and xs the steady-state solution. Show that xc , the solution to the associated
homogeneous equation to 2, is the transient solution in this case. 4 (c) To find a particular solution use the method of undetermined coefficients. Plug xp (t) into 2 and determine what coefficients determine a solution. (d) Take xp (t) = A sin(ηt) + B cos(ηt) from the previous problem. Show that
A sin(ηt) + B cos(ηt) = D sin(ηt + φ), where D= p F
(ω2 − η2 )2 (e) Note that D is largest when
E(η) = (ω2 − η2 )2 + 4λ2 η2 = η4 + 2(2λ2 − ω2 )η2 + ω4
is smallest. Show the following:
(i) If 2λ2 > ω2 , then the minimum of E(η) occurs at η = √
0.
(ii) If 2λ2 < ω2 , then the minimum of E(η) occurs at η = ω2 − 2λ2 . + 4λ2 η2 .

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