How to determine the free term in the second order linear differential equation with constant coefficients, for example, the second derivative of Y + the first derivative of y = e ^ 2x The form of the special solution is x ^ KQ (x) e ^ UX. I just don't know what the Q (x) should be and how to determine it. Sometimes it's ax + B, but there's a. under the guidance of the above example, q (x) = a, why can't I take other values? Isn't Q (x) a value ax + B

The right side is actually P (x) e ^ (2x), P is a polynomial of X, only p = 1, which is a polynomial of degree 0. The form of the special solution depends on whether the exponent 2 of E is the root and multiplicity of the characteristic equation B ^ 2 + B = 0. In this problem, 2 is not the characteristic root, that is, multiplicity k = 0, so the special solution is set to multiply e ^ (2x), that is AE ^ (2x)

From the values of the first four terms of the sequence 1,1 + 2 + 1,1 + 2 + 3 + 2 + 1 ····, we infer the value of the nth term 1 + 2 + 3 + ··· + [n-1] + N + {n-1 ··· + 3 + 2 + 1

This is simple:
The sum of the first n natural numbers should be known: a (n) = (n + 1) n / 2
Your topic means the sum of the first n natural numbers and the sum of the first n-1 natural numbers
So:
a(n)+a(n-1) = [(n+1)n/2]+[n(n-1)/2]=(n+n^2+n^2-n)/2=n^2

lim（n→∞）∑（k=1,n）1/k（k+2）

1 / K (K + 2) = 0.5 (1 / K -- 1 / (K + 2)), so the sum of the first n terms of the series is
0.5(1--1/3+1/2--1/4+1/3--1/5+...+1/n--1/(n+2))
=0.5(1+1/2--1/(n+1)--1/(n+2)),
The last two terms tend to 0, so the sum of series is 0.5 (1 + 1 / 2) = 3 / 4

From the first four terms of sequence 1,1 + 2 + 1,1 + 2 + 3 + 2 + 1,1 + 2 + 3 + 4 + 3 + 2 + 1, we infer the value of the nth term
The value of the first four items is deduced, and the proof is given

Prove: the nth item 1 + 2 + 3 + +n+(n-1)+… +1＝1+2+3+… n+n+(n-1)+… +1=（1+2+3+… n）×2-n
So the value of the nth term is (1 + n) × n / 2 × N-N = n + n ^ 2-N = n ^ 2

Why is LIM (x tends to infinity) [(3 + x) / (2 + x)] ^ 2x, [(1 + 3 / x) ^ x] ^ 2 equal to 6?

2X / (2 + x) is OK, but 2 / (1 + 2 / x) is not

Calculation: (- 3 and 4 / 5) + 1 + (- 1 / 5) =?

-3

If f (x) is differentiable at x = 0 and f (a) is not equal to 0, the solution of LIM {f (a + 1 / x) / F (a)} ^ x tends to infinity. The following x is the x-square. The detailed process is given
Experts help

1、
F (a) ≠ 0 can be used as denominator
The original formula = LIM (x - > ∞) e ^ {XLN [f (a + 1 / x) / F (a)]}
=lim(x->∞) e^{x[lnf(a+ 1/x)-lnf(a)]}
=lim(x->∞) e^{lnf(a+ 1/x)-lnf(a)]/(1/x)}
Top 0 bottom 0 lobida
LIM (x - > ∞) e ^ [1 / F (a + 1 / x)]
=e^[1/f(a)]
2、
If the title is derivable at a
lim(x->∞) e^{lnf(a+ 1/x)-lnf(a)]/(1/x)}
At this stage
Let f (a) = LNF (a) a be a variable
F'(a)=f'(a)/f(a)
And f '(a) = LIM (1 / X - > 0) LNF (a + 1 / x) - LNF (a)] / (1 / x)
So the result is e ^ f '(a)
=e^[f'(a)/f(a)]

Calculation √ (5 + 2 √ 6) + √ (7-4 √ 3) - √ (6-4 √ 2) - four times √ (1 - √ 2) ^ 4=

√(5+2√6)+√(7-4√3)-√(6-4√2)-⁴√(1-√2)⁴
=√(√3+√2)²+√(2-√3)²-√(2-√2)²-|-1√2|
=√3+√2+2-√3-2+√2-√2+1
=√2+1

lim sin4x/tan2x x→0

When x goes to zero,
SiNx and TaNx are equivalent to X,
therefore
Original limit
=lim(x→0) 4x/2x
=2
If we don't know the equivalence, we use the law of lobita to derive the numerator denominator at the same time,
Original limit
=lim(x→0) (sin4x)' / (tan2x)'
=lim(x→0) 4cos4x / [2/(cos2x)^2]
=4/2
=2

Calculation (+ 6 3 / 5) + (- 5 2 / 3) + (- 4 2 / 5) + (- 2 1 / 7) + (- 1) + (- 1 1 1 / 7)

(+ 6 and 3 / 5) + (- 5 and 2 / 3) + (+ 4 and 2 / 5) + (+ 2 and 1 / 7) + (- 1) + (- 1 and 1 / 7)
=(+ 6 and 3 / 5) + (+ 4 and 2 / 5) + (+ 2 and 1 / 7) + (- 1) + (- 1 and 1 / 7) + (- 5 and 2 / 3)
=[(+ 6 and 3 / 5) + (+ 4 and 2 / 5)] + [(+ 2 and 1 / 7) + (- 1) + (- 1 and 1 / 7)] + (- 5 and 2 / 3)
=11 + 0 + (- 5 and 2 / 3)
=5 and 1 / 3