Find the special solution of the differential equation y "- 4Y '+ 3 = 0 satisfying the initial conditions y (0) = 1, y' (0) = 5

Find the special solution of the differential equation y "- 4Y '+ 3 = 0 satisfying the initial conditions y (0) = 1, y' (0) = 5

The characteristic equation of the homogeneous equation y '' - 4Y '+ 3 = 0 is R & sup2; - 4R + 3 = 0, then the characteristic root is R1 = 1, R2 = 3
The general solution of the homogeneous equation y '' - 4Y '+ 3 = 0 is y = C1E ^ x + c2e ^ (3x) (C1, C2 are integral constants)
∵y(0)=1,y’(0)=5
∴C1+C2=1,C1+3C2=5
==>C1=-1,C2=2
So the special solution satisfying the initial condition is y = 2E ^ (3x) - e ^ X

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If 1 ≤ A5 ≤ 4, 2 ≤ A6 ≤ 3, then the value range of S6 is___ ．

A5 = a1 + 4D, A6 = a1 + 5D, so 1 ≤ a1 + 4D ≤ 4, 2 ≤ a1 + 5D ≤ 3, S6 = 3 (a1 + A6) = 6A1 + 15d, 6A1 + 15d = 15 (a1 + 4D) - 9 (a1 + 5d), so - 12 ≤ S6 ≤ 42

Prove LIM (n - > ∞) (2 ^ n) * (n!) / (n ^ n) = 0 by necessary conditions of convergence

Compare the latter with the former: because {2 ^ (n + 1) (n + 1)! / (n + 1) ^ (n + 1)} / {2 ^ n (n)! / (n) ^ n = 2 / (1 + 1 / N) ^ n tends to 2 / E

It is known that the sum of the first n terms of the sequence {an} is Sn, and for any n ∈ n *, there is always Sn = 2an-n, let BN = log2 (an + 1)
(1) Verification: the sequence {an + 1} is an equal ratio sequence
(2) Find the general formula of sequence {an}, {BN}

Sn=2an - n,S(n-1)=2a(n-1) - (n-1)
Sn-S(n-1)=an=2an - 2 a(n-1)-1 ,2a(n-1)=an -1
an +1= 2 [a(n-1)+1],(an +1)/[a(n-1)+1] = 2
So (an + 1) is an equal ratio sequence with a common ratio of 2
a1=2a1-1,a1＝1,a1 +1=2
an +1= 2*2^(n-1) = 2^n an = 2^n -1
bn=log2(an +1) = log2(2^n) =n

LIM (n tends to infinity) (n power of 1 + n power of 2 + n power of 3 + n power of 4) 1 / N power =?

Answer: Lim [(1 ^ n + 2 ^ n + 3 ^ n + 4 ^ n)] ^ (1 / N) = Lim [4 ^ n * ((1 / 4) ^ n + (2 / 4) ^ n + (3 / 4) ^ n + 1] ^ (1 / N) = Lim [4 ^ n] ^ (1 / N) * Lim [(1 / 4) ^ n + (2 / 4) ^ n + (3 / 4) ^ n + 1)] ^ (1 / N) = 4 * Lim [(1 / 4) ^ n + (2 / 4) ^ n + (3 / 4) ^ n + 1]

Given the equation (log2 x) ^ 2 + (T-2) log2 x + 1-t-m = 0, if t changes in the interval [- 2,2], the value of M is always positive, and the value range of X is obtained

Let P = f (T) = (log2x-1) t + (log2x) 2-2log2x + 1 ∵ P = f (T) be a straight line in the top rectangular coordinate system, so when t changes in the interval [- 2,2], the necessary and sufficient condition for P to be positive is f (- 2) > 0, f (2) > 0, that is: (log2x) 2-4log2x + 3 > 0, (log2x) 2-1 > 0

Given that the series ∑ is infinite above and converges below n = 1, UN, then Lim n-infinite, UN =? D = {(x, y) / upper line is 1

The necessary condition of series convergence is that the general term must tend to zero, which is the theorem of series
Therefore: Lim [n →∞] UN = 0
∫∫ 1 dxdy
=∫[0→2] 1dy∫[1→4] 1 dx
=2×3
=6

Let P = (log2x) 2 + (T-2) log2x-t + 1, if t changes in the interval [- 2, 2], P is always positive, try to find the value range of X

When p = (log2x-1) t + (log2x) 2-2log2x + 1, ∵ t ∈ [- 2, 2], P is always positive, ∵ - 2 (log2x-1) + (log2x) 2-2log2x + 1 ＞ 02 (log2x-1) + (log2x) 2-2log2x + 1 ＞ 0, the solution is log2x ＜ - 1 or log2x ＞ 3, that is, 0 ＜ x ＜ 12 or X ＞ 8

Let the positive series ∑ (n = 1 →∞) UN converge, and C be a constant, then the number in the following options must converge
A. Σ (n = 1 →∞) (root sign UN) B, Σ (n = 1 →∞) (UN + C)
C、∑（n=1→∞）（Un+C）² D、∑（n=1→∞）（Un²)
The answer is d

If ∑ UN converges, then the general term UN tends to 0 (when n tends to infinity) from the necessity of convergence

Given that the first n terms and Sn of sequence {an} satisfy log2 (Sn + 1) = n + 1, the general term formula of sequence {an} is obtained

From the known Sn + 1 = 2N-1, we get Sn = 2n + 1-1, so when n = 1, A1 = S1 = 3; when n ≥ 2, an = sn-sn-1 = 2n, and A1 = 3 does not conform to an = 2n, so the answer is an = 3 (n = 1) 2n (n ≥ 2)