Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm.

Get a Professional Answer

Via email, text message, or notification as you wait on our site. Ask follow up questions if you need to.

100% Satisfaction Guarantee

Rate the answer you receive.

Ask Dr Arthur Rubin Your Own Question

Dr Arthur Rubin, Doctoral Degree

Category: Calculus and Above

Satisfied Customers: 1493

Experience: Ph.D. in Mathematics, 1978, from the California Institute of Technology, over 20 published papers

13714381

Type Your Calculus and Above Question Here...

Dr Arthur Rubin is online now

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.

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

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)).

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.

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

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.