∂ ^ 4Y / & # 8706; X ^ 4 + & # 8706; ^ 2Y / & # 8706; T ^ 2 = 0 how to decompose into ordinary differential equation, the basic solution is y (x, t) = w (x) Q (T)

1. The form of fundamental solution is substituted into the equation;
2. Move one of the items to the right of the equal sign;
3. Divide the two sides by w (x) Q (T);
4. Let the left and right sides be equal to the constant P (unknown parameter);
5. Two ordinary differential equations about X and t can be obtained

Find the value of M. n, so that x + y = n, x = 2, y = 3, x + 2Y = m have the same solution
X + y = n, x = 2 is a system of equations, y = 3, x + 2Y = m is a system of equations

From x + y = n, x = 2, we can get 2 + y = n, then y = n-2
If two systems of equations have the same solution, then the above y is the same as that of the second system of equations, that is, y = 3, then y = n-2 = 3, n = 2 + 3 = 5
Similarly, if y = 3 is taken into x + 2Y = m, x = M-6 is obtained. Because the same solution as the first equation, then x = 2 = M-6, so m = 2 + 6 = 8
So the result is n = 5, M = 8

(y ^ 2-xy) y '+ 2Y = 0

Obviously, y = 0 is a solution
In addition, yes
(y-x)y'+2=0
Let z = Y-X be
zz'+z+2=0
Separation of variables
z-2ln|z+2|+x=C
Namely
y-ln(y-x+2)^2=C

Find the limit LIM (x tends to 0) x ^ n / e ^ ax (a > 0, n is a positive integer)

The molecule tends to zero
The denominator tends to e ^ 0 = 1
So the original formula = 0

It is known that the intersection point of line L1: y = - x + m and line L2: y = 2x-6 is in the fourth quadrant to find the value range of M

The intersection of line L1: y = - x + m and line L2: y = 2x-6 is (x, y) ① - x + M = 2x-63x = m + 6 ∵ the intersection is in the fourth quadrant ∵ x > 0 m + 6 > 0 ∵ m ∵ 6 ∵ ② y = - x + m x = m-yy = 2x-6, x = (1 / 2) y + 3m-y = (1 / 2) y + 3 (2 / 3) y = M-3 ∵ the intersection is in the fourth quadrant ∵ y < 0 M-3 < 0 ∵ m < 3

Find LIM (x tends to a) x [n] - a [n] / x-a, where n is a positive integer and the bracket represents the limit of n power!

You forgot to put brackets on both sides of the division sign / it's easy to produce ambiguity
You should be asking that
lim(x^n-a^n)/(x-a)
(the ^ sign is used to refer to the index in the computer)
Here is the factorization of x ^ n-a ^ n. if you are not familiar with it, find the old book and make it up
x^n-a^n = (x-a)[x^(n-1)+x^(n-2)a+...+xa^(n-2)+a^(n-1)]
In the formula, the 0 molecule x-a is removed, and the calculation of limit is directly replaced by the value tending to when it is reasonable. Here, X is replaced by a, so the original formula is equal to [x ^ (n-1) + x ^ (n-2) a +... + XA ^ (n-2) + A ^ (n-1)] = n * a ^ (n-1)

If the point P (m, 2m + 1) is in the third quadrant, then the value range of M is a, m < 0 B, m < - 0.5 C, M > - 0.5 D, - 0.5 < m < 0.5 D
Please write down the process in detail. Thank you

In the third quadrant, then M

This is a common problem of finding the limit of the reduced definite integral. The specific examples are as follows: n Lim1 / N ∑ sin (K π / N) = s [0,1] sin
My question is: if K π / N is taken as X, then the integrand function is SiNx, and the integral interval becomes [0, π], then how to find it? I tried this transformation method, and the result is not right? The above result is 2 / π, but according to the transformation method I said, the result is 2. I'm a graduate student, and I want to follow it to the end. What's wrong with me?

The example of the landlord is well written!
We divide the interval into n cells, each of which is 1 / n
The limit of Lim1 / N ∑ sin (K π / N) is equal to the area, in which the variable should be K instead of N. the value of n only affects the accuracy of finding the limit product of the surface, so n is regarded as a constant. Because the figure is determined, and the value of N affects the value range of K, so we regard K / N as a variable, and the value range of K / N is 1 / N ~ n / N, that is, 0 ~ 1 K π / N = x is equivalent to u π = x and u belongs to 0 ~ 1
It's equivalent to changing the variable to X and X belongs to 0 ~ π
If we want to write the integral model, it is the definite integral of ∫ sin (π U) Du in 0 ~ 1 or the definite integral of ∫ sin (x) d (x / π) in 0 ~ π
What this guy said upstairs is that if the integral interval is changed, the unit of the integral interval should also be changed correspondingly. What I said should be very specific

If | m-5 | = - (m-5), then the value range of M is

If | m-5 | = - (m-5), then the value range of M is
m-5

(1) Calculation + 2 * 2! + 3 * 3! + +N * n! (2) proof: K / (K + 1)! = 1 / K! - 1 / (K + 1)! (3) proof: 1 / 2! + 2 / 3! + n / (n + 1)
(1) Calculation + 2 * 2! + 3 * 3! + +n*n!
(2) Verification: K / (K + 1)! = 1 / K! - 1 / (K + 1)!
(3) Verification: 1 / 2! + 2 / 3! + n / (n + 1)! = 1-1 / (n + 1)

(1) AK = k * k! = (K + 1-1) k! = (K + 1)! - K! 1! + 2 * 2! + 3 * 3! +... + n * n! = 1! + 3! + 2! + 4! - 3! +... + (n + 1)! - n! = (n + 1)! - 1 (2) 1 / K! - 1 / (K + 1)! = 1 / K! - 1 / [(K + 1) k!] = (1 / K!) [1-1 / (K + 1)] = K / (K + 1)! (3) the length is limited, so the second conclusion is suggested

∂ ^ 4Y / & # 8706; X ^ 4 + & # 8706; ^ 2Y / & # 8706; T ^ 2 = 0 how to decompose into ordinary differential equation, the basic solution is y (x, t) = w (x) Q (T)