Um my PC broke for a while.. looks like I've a lot to answer.
@changsho:
Find two linearly independent power series solutions for the following differential equation about the ordinary point x=0
y''-xy'-(x+2)y=0
Basically you're searching for a function that looks like y=sum of an.x^n, n>=0
when you write that,
y''=sum of (n+2)(n+1).a(n+2).x^n, n>=0
xy'=sum of n.an.x^n, n>=1
xy=sum of a(n-1).x^n, n>=1
and 2y=sum of 2.an.x^n, n>=0.
So you're trying to find the numbers (an),n>=0 such that :
2.a2-2.a0=0, which means a2=a0,
and for n>=1, (n+2)(n+1).a(n+2)-(n+2).an-a(n-1)=0,
so a(n+2)=an/(n+1) + a(n-1)/(n+1)(n+2).
So 2 linearly independant solutions are the one with a0=0 and a1=1 and the one with a0=1 and a1=0, and then just hope that you don't need to get explicit terms.
@Mennen728:
A motorcycle cop is parked on the side of a highway reading a magazine when a woman driving a Ferrari passes him at 25 m/s. After a few attempts to get his motorcycle started the officer roars of 2 seconds later. At what average rate must he accelerate if 30.5 m/s is his top speed and he is going to catch her just at the state line 2000 meters away.
This one is not too difficuly. Say Vf = speed of the ferrari, Vc = max speed of cop, t= 2s,the delay before the cop starts, and a = cop acceleration.
First, we need to know the distance between the cop and the ferrari after the acceleration phase.
That phase lasts for T1=Vc/a, the Ferrari will have travelled for T1.Vf=Vf.Vc/a during that time, and the cop for 1/2.(a.T1²)=1/2.Vc²/a
The distance between the 2 after that phase is D=Vf.(t+T1)-1/2.Vc²/a
So D=1/a.[Vf.Vc-1/2.Vc²] + Vf.t
From that point, they travel at constant speed, so the cop meets the Ferrari after the time T2=D/(Vc-Vf) = 1/a.(things) + (things)
The goal is to stop the ferrari at the distance L=2000 m.
By the moment they meet they'll have travelled by (t+T1+T2).Vf
So L=(t+T1+T2).Vf = … = 1/a.(things) + (things)
So by putting all these equations together, you eventually get an affine equation on a, which is very simple to solve.
@Ral:
2.) At a certain location, the horizontal component of the earth's magnetic field is 2.5 x 10^-3 T, due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.
It's very basic. Take formulas that give the amplitude of the weight and the magnetic force : P = m.g and FLorentz = B.speed.(some constant which I don't remember but that you should have somewhere)
Say that these are equal, and solve : speed = m.g/(B.(…)
sanji37, are you serious ? Take a calculator and do the operations =/.