Recent content by ascheras

sorry, i meant (2,1).

ok, so I've never done a problem like this one before: find all solutions: 24x + 11y == 4 (mod 35) 5x + 7y == -13 (mod 35). This reduces to: 24x + 11y == 4 (mod 5) 5x + 7y == -13 (mod 5). and 24x + 11y == 4 (mod 7) 5x + 7y == -13 (mod 7). Solving the two, i get (2,1) and...

ok, upon doing it (mod 7) and (mod 5), i got (3,4) (mod 7) and (1,2) (mod 7). does that sit well? or should i now apply the CRT?

maybe i don't have the right zeros... i'll try again and see what i get.

I'm having problems finding all integer solutions to some of the higher degree polynomials. for p(x)= x^3− 3x^2+ 27 ≡ 0 (mod 1125), i get that 1125 = (3^2)(5^3). p(x) ≡ 0 (mod 3^2), p(x) ≡ 0 (mod 5^3). x ≡ 0, 3, 6 (mod 3^2) for 3^2 for 5^3, x ≡ 51 (mod 5^3) then i get x=801, 51, 426 (mod...

Show that the diophantine equation x^2 - y^2= n is solvable in integers iff n is odd or 4 divides n.

Let n= (2^r)*q, where q has no prime divisors. Then you can write (2^n) +1= (y^q) +1, where y=(2^r). Then (2^n) +1= (y+1)(y^(q-1)- y^(q-2) +...+ (y^2) - y+1). Here, y+1= (2^r)+1 >1 and there are two factors. This (2^n)+1 cannot be prime if the other factor is also > 1. That happens unless...

Prove the following is valid only when n is an odd integer. x^n + 1= (x+1)(x^n - X^n-2 + ... + (x^2 - x + 1). It's an easy 3 line proof.

here's some to try (i'll have answers either on thursday or tuesday): f(x)= sin(x^(x^1/2) + (pi*log x)), x>0 f(x)= log(log(log(x^2 + 16))), x>0 f(x)= integral (from 0 to x) of exp((x-y)^1/2) dx

i have the answers in the back of my book. the problem is that i don't understand how to get the wronskian at t=0. this is a variation of parameters of higher order DE (cookiemonster). Y(t) is the general solution to the homogeneous part of the problem.

Ha, i wish i was taught in class. i get u1e^t + u2cost + u3sint = Y(t) the i get the 3X3 system of equations with the final row equalling e^(-t) sint. I just don't know what to do with it.

Use the method of variation of parameters to determine the general solution of the given DE: y''' - y" + y' -y= e^(-t)sint

reply i'm glad you decided to help, but your belittling was unwelcome. i figured this problem out already using the variation of parameters. thank you anyway. and i did fully understand the problem and it was stated that way. it is understood that y1 is the solution to the homogenous equation...

I can't seem to properly solve this problem: ty" - (1+t) y' + y= t^2 e^(2t) , t> 0; ysub1= 1+t any help would be appreciated