Solution, dy / DX + y = x, y (0) = 1 The answer is y = X-1 + 2E ^ - X Laplace transformation is used, and the process will not change

Solution, dy / DX + y = x, y (0) = 1 The answer is y = X-1 + 2E ^ - X Laplace transformation is used, and the process will not change

Let u = Y-X, Du / DX = dy / DX-1
The original equation is reduced to Du / DX = - U-1 = > Du / (U-1) = - 1) DX
Integral: ln | U + 1 | = - x + C = > u = - 1 + C1 e ^ (- x)
y = x - 1 +C1 e^(-x)
Substituting y (0) = 1, = > C1 = 2
y= x -1 + 2 e^(-x)

The density of cement is 3.06, and the ratio of cement slurry to sodium silicate is 1:0.1. What is the dosage of sodium silicate? The ratio of cement slurry is 1:1, and the dosage per cubic meter is 754kg

When the cement slurry ratio is 1:1, the density of cement is 3060 (kg / m ᦉ 179;). The specific gravity of water is 1000 (kg / m ᦉ 179;). Then it becomes 2 cubic meters of cement slurry, and the specific gravity is: (1000 + 3060) / 2 = 2030 (kg / m ᦉ 179;). Then 1 cubic meter of cement slurry plus 2030 * 0.1 = 203 kg of sodium silicate

In the sequence {an}, an = 2A (n-1) + 2 ^ n + 1, A3 = 27 to find the general term formula of an?

Undetermined coefficient method
Firstly, A3 = 27, A2 = 9, A1 = 2 are obtained
Let an integer λ. We can construct a {(an + λ) / (2 ^ n)} sequence first
(an+λ)/(2^n) - (a(n-1)+λ)/(2^(n-1)) = (an+λ-2a(n-1)-2λ)/(2^n)
By substituting an = 2A (n-1) + 2 ^ n + 1 into the above formula, we can get
(an+λ)/(2^n) - (a(n-1)+λ)/(2^(n-1) = 1 + (1-λ)/(2^n)
Therefore, let λ = 1
So (an + 1) / (2 ^ n) - (a (n-1) + 1) / (2 ^ (n-1) = 1
That is, the sequence {(an + λ) / (2 ^ n)} is an arithmetic sequence with 1 as the tolerance and (a1 + 1) / 2 = 3 / 2 as the first term
The general term is: (an + λ) / (2 ^ n) = 2 / 3 + n-1 = 1 / 2 + n
an=(1/2+n)2^n -1
Substituting a 1, a 2, a 3 for checking calculation also conforms to the general formula
I suggest you write on the paper. Maybe it's not very clear

Given that the function f (x) = | x ^ 2-2x-3 | - a satisfies the following conditions respectively, find the value range of real number a_______
(2) Function has three zeros; (3) function has four zeros
(2) When a = 4, the image has three zeros
(3) What about it?
It should be 0

Please refer to the picture above
The function has three zeros: a = 4
The function has four zeros: 0 < a < 4

987 + 402 = () (simple calculation:

987+402=900+400+87+2=1389

The diameter of the bottom surface of a cylinder is 4 cm. Cut along the diameter of the bottom surface of the cylinder and divide it into two identical three-dimensional figures. The surface area increases by 32 square centimeters. What is the original surface area of the cylinder?

The increased surface area is the size of two rectangles; length = cylinder height, width = cylinder diameter, so: cylinder height = 32 ／ 2 ／ 4 = 4cm, radius = 4 ／ 2 = 2cm, surface area = 2 × 3.14 × 2 & # 178; + 2 × 3.14 × 2 × 4 = 75.36cm2 ~ a moment forever 523 for you, I wish you progress in learning ~ ~ ~ ~ if you agree

Factorization factor 2x ^ 3-4x ^ 2Y ^ 3 + 6x ^ 2Y ^ 2

2x^3-4x^2y^3+6x^2y^2
=2x^2(x-2y^3+3y^2)

X ^ 2 + y ^ 2 + x-6y + M = 0 and X + 2y-3 = 0 intersect at p q, Op perpendicular to p q, find M

In this paper, the standard formula of the circle square is: [x + (1 / 2)] ^ 2 + (Y-3) ^ 2 = (37-4m) / 4 (1) So, the center coordinate of the circle is (- 1 / 2,3) and the simultaneous line and circle equation get: x ^ 2 + X + y ^ 2-6y + M = 0, x + 2y-3 = 0 = = > (2y-3) ^ 2 - (2y-3) + y ^ 2-6y + M = 0 = = = > 4Y ^ 2-12y

The absolute value of the quadratic power of x minus 1 is greater than 2m plus 1

|x²-1|>2m+1
If 2m + 10, M > - 1 / 2
Then x & sup2; - 12m + 1
x²2m+2
-√(-2m)

Given that the function f (x) = 2x + 3, the sequence {an} satisfies A1 = 1, and a (n + 1) = f (an), then the general term formula of the sequence is - (all following a are subscripts)

A (n + 1) = f (an), that is, f (an) = 2An + 3 = a (n + 1)
2（An +3)=An+1 + 3
That is, an + 3 is an equal ratio sequence with the common ratio of 2 and the first term of 4
An + 3 = 4 * 2 ^ n-1, that is, an = 4 * 2 ^ n-1 - 3