Find the complete homogeneous solution using variation of parameters
The solution is
Therefore, and
Plugging this back into the original homogeneous equation,
so
Checking the answer
Plugging this into the original homogeneous equation
Plugging in values for y and its derivatives, everything cancels out to zero.
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Find and plot the solution for the L2-ODE_CC
This is a linear, first order ODE with constant coefficients.
To find the general solution to this ODE set
so that and
Substituting in y to the ODE and factoring out we get:
Using the quadratic formula to solve for r we get
where and
Solving to get
Since we have a repeated root, we need to find v(x) so that y2(x) = v(x)y1(x)
Taking the first and second derivative of y2(x) we get:
Substituting into the original ODE, we get:
solving for so v(x) = kx + c
So y2(x) = x y1(x)
We get the general solution
Now with the initial values y(0) = 1 and y'(0) = 0
,
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
How does the frequency of the harmonic oscillation change if we (i) double the mass (ii) take a spring of twice the modulus?
Could you make a harmonic oscillation move faster by giving the body a greater push?
Now double the mass
The frequency is decreased by .
Multiply k by 2
The frequency is increased by .
No because frequency depends on the ratio of the spring modulus and mass.
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Show the maxima of an underdamped motion occur at equidistant t-values and find the distance.
Determine the values of t corresponding to the maxima and minima of the oscillation . Check your result by graphing y(t).
The general solution of underdamped motion is
The maximas occur at
Set the two equations equal to each other a solve for t.
where n=0,1,2,3.....
shows delta is a constant.
The periodic distance between maximas is
To find critical points, set y'(t)=0
where n=0,1,2,3...
As seen in the graph, the maximum of t was at and the minimum was at .
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Using the formula for Taylor series at x = 0 (the origin, i.e., McLaurin series), develop into Taylor series at the origin x = 0 for the following functions: cos x, sin x, exp(x), tan x, and write these series in compact form with the summation sign and a single summand.
Part 1
cos x
Taylor Series:
when
n = 0, 1, 2, 3 ... N
2n = 0, 2, 4, 6, ... 2N
Part 2
sin x
Taylor Series:
when
Part 3
exp x
Taylor Series:
when
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Part 1
Find and plot the solution for the L2-ODE-CC
Initial conditions: y(0) = 1, y'(0) = 0
No excitation: r(x) = 0
Part 2
In another Fig., superpose 3 Figs.: (a) this Fig.,
(b) the Fig. in R2.6 P. 5-6, (c) the Fig. in R2.1 P. 3-7.
Part 1
The characteristic equation of the given ODE is:
Using the quadratic formula to solve for
where
Solving to get
,
Therefore, the general solution of the given ODE is
Now we solve for , using the given initial conditions
We have
Substituting into ,
We get,
We have
Differentiating , we get:
.
Substituting into , we get:
Therefore, we get
Hence the solution of the given ODE is
Fig. 1:
Part 2
Fig. 2:
Fig. 3:
Fig. 4:
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Consider the same system as in the Example p.7-3, i.e.,
the same L2-ODE-CC (4) p.5-5 and initial condi-
(2) p.3-4, but with the following excitation:
Replacing with and after simplifying we get,
The root here is . So we can solve for our constants,
Using the initial conditions y(0)=4 and y'(0)=-5 we can solve for our constants,
So the solution is,
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Plot the error between the exact derivative and the approximate derivative, i.e.
from
For ε = .0001, .0003, .0006, and .001 and λ =.3
Since the above equation is the error between the exact derivative and the approximate derivative, it must be plotted with the correct values of \epsilon and \lambda, from x = -15 to 15
ε=.0001
ε=.0003
ε=.0006
ε=.001
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Find the complete solution for , with the initial conditions
,
plot the solution y(x)
Particular Solution
Homogeneous Solution
Initial conditions
General Solution
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.