Differential equation: using substitution method to solve differential equation DX / dt + TX ^ 3 + X / T = 0

Differential equation: using substitution method to solve differential equation DX / dt + TX ^ 3 + X / T = 0

Let z = 1 / X & # be substituted into the original equation
dz/dt-2z/t=2t.(1)
∵ equation (1) is a first order linear differential equation
From the general solution formula, it is concluded that
The general solution of equation (1) is Z = T & # 178; (c + 2ln │ t │) (C is an integral constant)
==>1/x²=t²(C+2ln│t│)
==>x²t²(C+2ln│t│)=1
Therefore, the general solution of the original equation is X & # 178; T & # 178; (c + 2ln │ t │) = 1 (C is an integral constant)

D ^ (2) x / dt ^ 2 + x = t + e ^ t to find the general solution of differential equation

The general solution of homogeneous equation d ^ (2) x / dt ^ 2 + x = 0 is C1 (Sint) + C2 (cost) because s ^ 2 + 1 = 0, s = I or S = - I. now it is very obvious to find a special solution of D ^ (2) x / dt ^ 2 + x = t + e ^ T. there is a special solution x = t + (1 / 2) e ^ t, so the general solution of differential equation d ^ (2) x / dt ^ 2 + x = t + e ^ t is t + (...)

lim x→n (√n+1-√n)*√(n+1/2)
lim x n→∞ (√n+1-√n)*√(n+1/2)

lim x n→∞ (√n+1-√n)*√(n+1/2)
Multiply by (√ n + 1 + √ n) and divide by (√ n + 1 + √ n)
Lim x n →∞ (√ n + 1 - √ n) * √ (n + 1 / 2) = Lim x n →∞√ (n + 1 / 2) / (√ n + 1 + √ n), the numerator and denominator are divided by √ n
The original formula = Lim x n →∞ √ (1 + 1 / 2n) / (√ n + 1 / N + 1) = 1 / (1 + 1) = 1 / 2

Let f (x) = INX PX + 1, prove: LN2 ^ 2 / 2 ^ 2 + Ln3 ^ 2 / 3 ^ 2 + +lnn^2/n^2

It is proved that P = 1
f(x)=lnx-x+1,x>=1
f'(x)=(1-x)/x1
Then f (x) decreases monotonically on x > 1, and f (x) can be continuous on x = 1
f(x)1,lnx-x+11
Lnx1
We take n & sup2; (> 1) to replace the above formula x, where
lnn²

lim(n->∞) n^2 [x^(1/n)-x^(1/n+1)]
The answer is LNX
No mistake. The original problem is x to the power of N, X to the power of N + 1,
Do not understand

Ha ha
Think of X as a constant
1 / N as x1, 1 / (n + 1) as x2
n^2 [x^(1/n)-x^(1/n+1)]=[(n+1)/n][1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]
[1 / n-1 / (n + 1)] * [x ^ (1 / N) - x ^ (1 / N + 1)] is the derivative of a ^ x at 0
The answer is LNX

If a new operation "△" is specified, that is, m △ n = m + 2n, for example, 3 △ 5 = 3 + 2 × 5 = 13, then what is the value of X in 4 △ (2x + 1) = x?

∵ m △ n = m + 2n, ∵ 4 △ (2x + 1) = 4 + 2 × (2x + 1) = 6 + 4x; ∵ 4 △ (2x + 1) = x, ∵ 6 + 4x = x, that is, 3x = - 6. Divide both sides of the equation by 3 at the same time, and the value of X in x = - 2, ∵ 4 △ (2x + 1) = x is - 2

Find Lim [x ^ (n + 1) - (n + 1) x + n] / (x-1) ^ 2 x -- > 1
=lim(t->0) [ [ 1 + (n+1)t + (n+1)n/2t^2 + o(t^2)] -(n+1)-(n+1)t + n]/t^2？ Don't you understand?

① Let: x = 1 + T (T - > 0) LIM (x - > 1) [x ^ (n + 1) - (n + 1) x + n] / (x-1) ^ 2 = LIM (T - > 0) [(1 + T) ^ (n + 1) - (n + 1) (1 + T) + n] / T ^ 2 [binomial expansion, O (T ^ 2) denotes the infinitesimal of higher order of T ^ 2: (n + 1) n (n-1) T ^ 3 / 3! +... + T ^ (n + 1)] = LIM (T - > 0) [...]

If (x ^ m ／ x ^ 2n) ^ 3 ／ x ^ M-N and - 1 / 4x ^ 2 are of the same kind, find 4m ^ 2-2n ^ 2 (+ ﹏ +) ~ halo

（x^m÷x^2n)^3÷x^(m-n)=x^3(m-2n)÷x^(m-n)=x^(3m-6n-m+n)=x^(2m-5n)
It is similar to - 1 / 4x ^ 2
Then 2m-5n = 2, 2m = 5N + 2
4m^2-2n^2=(5n+2)^2-2n^2=23n^2-20n+4=23(n-10/23)^2-8/23
Unable to find the final result, there is a minimum value of - 8 / 23

Finding the limit of X to 0 in (xsin1 / x x x / SiNx x x / c0sx)

When x goes to zero,
Sin1 / X is a bounded function, then x * sin1 / X tends to 0
From the important limit, we can know that X / SiNx tends to 1,
Cosx tends to 1, so x / cosx also tends to 0
So get
Original limit = 0 + 1 + 0 = 1

If (z-x) &# 178; - 4 (X-Y) (Y-Z) = 0, it is proved that x + Z = 2Y has learned 15.2 multiplication formula
There is more than one solution

(z-x)²-4(x-y)(y-z)
=z^2-2xz+x^2-4(xy-xz-y^2+yz)
=x^2+z^2-2xz-4xy+4xz+4y^2-4yz
=x^2+z^2+2xz-4xy+4y^2-4yz
=(x^2+2xz+z^2)-(4xy+4xz)+4y^2
=(x+z)^2-4y(x+z)+4y^2
=(x+z-2y)^2=0
x+z-2y=0
x+z=2y