The difference between the second order differential equation with reduced order and the second order linear differential equation with constant coefficients How can these two types of equations be distinguished I'm always confused about several types of differential equations I have a terrible headache

@Second order differential equation of reducible order 1, y '' = f (x) type. The characteristic of this kind of equation is that the right end of the equation only contains independent variable x, and the general solution of the equation can be obtained by integrating twice. 2, the characteristic of this kind of equation is that the right end of the equation does not contain unknown function y

The differential equation f '(x) = 1 + (f (x)) ^ 2 is expressed as F "(x) by the formula containing F (x)

∫ (lower limit 1, upper limit x) TF (T) DT = XF (x) + x ^ 2 the solution of F (x) differential equation

Because of [1, x] t (T) DT = XF (x) 2 (1) in this paper, we have obtained the following x (x) x (x) from the X (x) we have: XF (x) (x) is the X (x) (x (x) (x) (x) (x) we (x (x) is the X (x (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) we (x) (x) (x (x) = f (x (x) f (x (x (x) = x (x (x (x) + (x) + (x) + (x) + (x) + (x) + (2x (x) + (1-x) / x) x (x) x (x) x (x) f (x) x) x (x (x) f (x) x (x (x) in this [8747474747474747474747474747474747474747474747474747 [2 [2 [2 [2, x [2 [2 [8747xde ^ (- x)

Skillfully calculate 1 / 1 × 2 + 1 / 2 × 3 + 1 / 3 × 4 + 1/49×50

1/1×2+1/2×3+1/3×4+… 1/49×50
=1-1/2+1/2-1/3+… +1/49-1/50
=1-1/50
=49/50

Why is the limit of X divided by TaNx equal to 1 when x approaches 0

Because the numerator and denominator tend to 0, so it's 0 / 0. You can use the l'Hopital rule
Lim = x '/ (TaNx)' = 1 / sec ^ 2 x (evaluated at x = 0) = 1 / 1 = 1

How to calculate (1 / 2) + (1 / 2 + 1 / 3) + (1 / 4 + 2 / 4 + 3 / 4) +... + (1 / 50 + 2 / 50 + 3 / 20 +... + 48 / 50 + 49 / 50)?
I'll use it today. As long as you answer quickly and accurately, I'll choose you as the best answer immediately. I'll use the division sign △ instead of /, which is easy to mix up
For a natural number n, if we can find the natural numbers a and B such that n = a + B + AB, then n is a good number. For example: 3 = 1 * = + 1 + 1 * 1, then 3 is a good number. How many good numbers are there from 1 to 20?

/2+(1/3+2/3)+（1/4+2/4+3/4)+...+(1/50+2/50+3/50+...+48/50+49/50)
The original solution is 1 / 2 + 1 + 3 / 2 + 2 +... + 50 / 2
=1/2+2/2+3/2+...+50/2
=（1+2+… +50）/2
=1275/2
Use 1 + 2 + +N = (1 + n) n / 2

Can the limit of tan3x / TaNx be equal to 0 when x tends to π / 2?
For example, I worked out that the answer is 0 and 1 / 3 (I got two results by solving quadratic equation with one variable in some magical way), but the answer is 1 / 3, that is to say, the limit can't be 0, why?

tan3x/tanx = (sin3x / sinx ) * (cosx / cos3x)
lim (cosx / cos3x) = lim -sinx / [ (-3) sin3x] = -1/3
lim (sin3x / sinx ) = -1
The original formula is 1 / 3
If limit exists, it must be unique

Solution equation: x + 3x + 4 − x + 4x + 5 = x + 1x + 2 − x + 2x + 3

The original equation can be transformed into 1-1 x + 4-1 + 1 x + 5 = 1-1 x + 2-1 + 1 x + 3, which can be reduced to 1 x + 4-1 x + 5 = 1 x + 2-1 x + 3, 1 (x + 4) (x + 5) = 1 (x + 2) (x + 3) (x + 4) (x + 5) by multiplying both sides of the equation to get (x + 2) (x + 3) (x + 4) (x + 5), and the solution is x = - 72

lim┬(n→∞) (2^x-1)/(2^x+1)

Let a = 2 ^ X
If x →∞, then a →∞
So 1 / a → 0
Original formula = (A-1) / (a + 1)
Divide up and down by a
=(1-1/a)/(1+1/a)
So limit = (1-0) / (1 + 0) = 1

Calculation: 1: (- x ^ n) ^ 2 + (x ^ 2) ^ N-X ^ n * x ^ 2 2: [(- n) ^ 2] ^ 5 / [(- n * n ^ 3) ^ 2 * 2n ^ 2]
:3:-a*a^5-(a^2)^3-(-2a^3)^2

1：(-x^n)^2+(x^2)^n-x^n*x^2 =x^2n+x^2n-x^(n+2)=2x^2n-x^(n+2)2：[(-n)^2]^5/[(-n*n^3)^2*2n^2]=[n^2]^5/[(-n^4)^2*2n^2]=[n^10]/[(n^8)*2n^2]=[n^10]/[2n^10]=1/2:3:-a*a^5-(a^2)^3-(-2a^3)^2=-a^6-(a^6)-(4a^6)=...