General solution of differential equation y "- 6y '- 16y = 0

General solution of differential equation y "- 6y '- 16y = 0

The characteristic equation is R ^ 2-6r-16 = 0, r = 8, - 2
So y = C1E ^ (8x) + C1E ^ (- 2x)

Ordinary differential equation y '= x3y3 XY by proper transformation

The equation y '= x ^ 3Y ^ 3-xy is the Bernoulli equation, divided by Y ^ 3:
y'/y^3=-x/y^2+x^3
U = 1 / y ^ 2 u '= - 2Y' / y ^ 3 substituting:
U '= 2ux-2x ^ 3, which is a first-order linear differential equation
1/y^2=u=e^(x^2)(C+∫[-2x^3e^(-x^2)]dx]
=Ce^(x^2)+x^2+1

Limn -- > ∞ (1 + 1 / 1 + 2 + 1 / 1 + 2 + 3 +... + 1 / 1 + 2 + 3 +... + n),

1/(1+2+…… +n)=1/[n(n+1)/2]=2/[n(n+1)]=2*[1/n-1/(n+1)]
So the formula in the limit = 2 * {(1 / 1-1 / 2) + (1 / 2-1 / 3) + +[1/n-1/(n+1)]]
=2*(1-1/(n+1)]
n→∞,1/(n+1)→0
So limit = 2 * (1-0) = 2

Given that point a (a + 3, 2-3a) is in the second quadrant, the value range of a is______ ．

∵ point a (a + 3, 2-3a) is in the second quadrant, ∵ a + 3 ＜ 0, ① 2 − 3a ＞ 0, ②, a ＜ - 3 is obtained by solving inequality ①, and a ＜ - 23 is obtained by solving inequality ②, so the value range of a is a ＜ - 3

How to find the limit limn approaching positive infinity (2 ^ n-3 ^ n) / 4 ^ n?

lim (2^n-3^n)/4^n
=lim (1/2)^n - lim (3/4)^n
=0-0, because 1 / 2

If point a (2x-4,6-2x) is in the second quadrant, then the value range of X is?

2X-4 is less than 0, X is less than 3, 6-2x is greater than 0, X is less than 2, so x is less than 2

Proving limn ^ 1 / N = 1 by limit definition

It is proved that limn ^ (1 / N) = 1: n ^ (1 / N) = 1 + HN, where HN > 0, and
　　　n = (1+hn)^n > C(n,2)(hn)^2 = [n(n-1)/2](hn)^2,
So, there is
　　　0

If a (2x-5, 6-2x) is in the fourth quadrant, find the value range of A______ ．

∵ a (2x-5, 6-2x) is in the fourth quadrant, ∵ 2x-5 > 06-2x < 0, the solution is x > 3

LIM (x - > 1) (x ^ (n) - 1) / x-1) =? N is a positive integer,

lim(x->1)(x^(n)-1)/x-1)=lim(x->1)(x^(n-1+x^(n-2)+…… +x+1)(x-1)/x-1)=lim(x->1)[x^(n-1)+x^(n-2)+…… +x+1]=n

If a (2x-5, 6-2x) is in the fourth quadrant, find the value range of A______ ．

∵ a (2x-5, 6-2x) is in the fourth quadrant, ∵ 2x-5 > 06-2x < 0, the solution is x > 3