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Dr Arthur Rubin
Dr Arthur Rubin, Doctoral Degree
Category: Calculus and Above
Satisfied Customers: 1561
Experience:  Ph.D. in Mathematics, 1978, from the California Institute of Technology, over 20 published papers
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Hi, here are some maths questions that need to be done with

This answer was rated:

Hi, here are some maths questions that need to be done with detailed working out that looks simple. There are six questions and I will upload them once there is an expert available. They are based on differential equations and sequences and series.

Thanks

Dr Arthur Rubin :

I'm willing to try.

Customer :

Yeah Thanks

Dr Arthur Rubin :

O

Dr Arthur Rubin :

(Sorry about that.) I recall differential equations fairly well.

Customer :

I have uploaded. Hope you have received.

Dr Arthur Rubin :

OK, I can do that. I should be able to upload a PDF solution in an hour or two.

Customer :

Sounds great. Thanks Arthur for helping again...

Dr Arthur Rubin :

OK, here should be a complete solution. Let me know if you need clarification or more detail. (As I'm on the Pacific coast of the US, I'm going to go to sleep shortly. I'll be back within 7 hours.)

Customer :

Hi Arthur,

Customer :

Hi Arthur,

Customer :

Hi Arthur, thanks let you know if I need clarification later when I look at it fully.

Customer :

Hi Arthur, Thanks let you know if need clarification when I look at it fully

Customer :

Sorry my computer is acting up

Dr Arthur Rubin and other Calculus and Above Specialists are ready to help you
Customer: replied 3 years ago.

Hi Arthur,


 


I was wondering for question 3 is it possible to get a more detailed working out on how the answer e^3x(cosx+Bsinx) was found. Thanks

Well, if you try substituting y = exp (t x) into the equation d^2y/dx^2 - 6 dy/dx + 10 y = 0,
you get t^2 - 6t + 10 = 0, which has roots t = 3 ± i. Hence, the general solution is

y = a exp( (3+i) x) + b exp((3-i)x) = exp(3 x) ( (a+b)cos(x) + i (a-b)sin(x)).

Does that help?

Customer: replied 3 years ago.

Where did the 3 come from... Is it a factor of 6

It's the quadratic equation; the solution of a t^2 - b t + c = 0 is
t = (b ± sqrt(b^2 - 4 a c))/(2 a). In this case, we get
t = (6 ± sqrt(36 - 40))/2 = (6 ± sqrt(-4))/2 = (6 ± 2 i)/2 = 3 ± i.

Customer: replied 3 years ago.

Yeah, now I am starting to understand...

Alternatively, t^2 - 6 t + 10 = (t-3)^2 + 1, so t^2 - 6 t + 10 = 0 converts to
(t-3)^2 = -1, or t-3 = ±i.
Customer: replied 3 years ago.

About question 4. Which formula was used to get the sequences for question a, b and c. And if it's possible could you please give me a list of all the sequences and series formula's generally used. Thanks

Sorry about the delay.

For 4 (a), it was just substituting values for "n", noting sqrt(n^3) = n sqrt(n).
For 4 (b), it was just substituting and adding, with no indication that there is a general formula for the partial sums, although the sum of the convergent series might be found as.
For 4 (c), the usual convergent series that can be used are:
sum n^-(1+ε),
sum (1/(n (log n)^(1+ε)))
sum(1/(n (log n) (log log n)^(1+ε))),
etc.

I'm heading off again; should be back in about 3 hours, at which time I'll probably need to get some sleep.