Recent content by Alphaboy2001

Hi Tiny-tim, you are one of the smartest guys here :D Anyway if look at my teachers definition of T1 and T2 its their inclusion which gives f(u,v) = 0 for any integer l and m, if and only if l and m are in Z. and what I am trying to do it to insert T1 + T2 and T1*T2 into f(u,v) and as...

Solution to diff. eqn as a ring(killing me - please look at my result!) Homework Statement I am presented with the following problem given the direction field f(u,v) = \left( \begin{array}{ccc} -sin(u) & sin(v) \\ cos(u) & cos(v) \end{array} \right) as \left( \begin{array}{c}u^{'} &...

Hi Tiny Tim, Its here that I am confused. Because if I set the RHS to zero, then I the homogenous solution from (2) and solution for inhomogenous system? If e^\int(b(s) ds) is a solution for the homogenous system. How do I find solution for the inhomogenous without having suitable b...

Hello again tim, and thanks for your answer. I meant the bound, sorry I am messed up. Anyway if I choose x = e^{\int b(t) \ ds} and insert this to obtain the expression (e^{\int b(t) \ ds})^2 -2b(e^{\int b(t) \ ds}) + (b^2 - b) \cdot (e^{\int b(t) \ ds}) = \ e^{-\int_{t_0}^{t_1}...

Hi again, If I set x t to be x = e^{\int b(s) \ ds} ?? what do I regarding the bound on rhs of inhomogenous equation?? since x(t_0) - x'(t_0) = 0 then x(t_0) - x'(t_0) = 0?? remove it?? Sincerely Alphaboy

But what about limit on b(t)?? Since the original (3) says find the solution for x''(t) - 2bx(t) + (b^2 - b'(t)) = e^{\int_{t_0}^{t} b(s) ds} then x(t_0) = x'(t_0) = 0 Then x_{p}(t) = e^{B(t)} - e^{B(t_0)} Is the solution here for (3) then x(t) = C \cdot e^{\int b(t) dt}...

What I don't get about question (2) Which says Find the general solution of (*) Isn't that in fact part of question(1) in the answer for question(2)? Is there a specific theorem here what I use to prove that (2: answer) Such that The generalized solution of (*) is supposedly x(t) =...

Hello again Tim, This is then the solution for x'(t) - bx = 0 don't I need to show here too that this is also the solution for x''(t) - 2bx'(t) + (b^2- b'(t))x = 0*?? Sorry for the stupid question now. Is this then done by using the fact that there is analogy between every b \in C^{1}...

I don't know if reason why we are talking past each other (I fear) is that my professor maybe is using his own reasoning (like talking about solutions for diff. eqn belong to their own respective sets??))) Anyway trying to solve x' = bx I get that x(t) = e^{b(t)} + C and since x \in...

Sorry Mr. Tiny Tim I made such a stupid error in the post above that I should be send to Guantanamo :sad: Anyway using Your calculations x'' - 2bx' + (b2 - b')x = bx' + b^2x - 2bx' + b^2x - bx' = 0 Putting this into order. 2bx^2 - 2bx' = 0 and since x' = bx Then...

Hello tiny-tim, I was referred to this forum by a friend and then I got stuck with the bigger problem which will follow below, I decided to check this forum out :) by the way there is typoo is my original equation: \frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0...

Homework Statement Hi I need to solve the following: \frac{d^{2} x}{dt} + 2 b \frac{dx}{dt} + (b^2 - \frac{db}{dt}) x = 0[/tex] 2. The attempt at a solution Isn't this equivalent to??:: [itex]x^{''} + 2bx^{'}+ (b^2 - b^{'})x = 0 \iff x^2 + 2x +1 = 0 If I let b = 1. which...