Differential equation 2Y '' - sin2y = 0, initial condition y (x = 0) = pi / 2, y '(x = 0) = 1!

Differential equation 2Y '' - sin2y = 0, initial condition y (x = 0) = pi / 2, y '(x = 0) = 1!

Let y '= P, then y' '= DP / DX = DP / dy * dy / DX = PDP / Dy, so the differential equation is 2pdp = sin2ydy, that is, D (P ^ 2) = -- 0.5d (cos2y), so p ^ 2 = -- 0.5cos2y + C. using the known conditions y (0) = pi / 2, y' (0) = P (0) ^ 2 = 1, we get 1 = -- 0.5cos (2 * pi / 2) + C, then C = 0.5. So p ^ 2 = 0.5 (1 -- cos2y) = sin ^

How can y + 1 = - 1 / 2 (X-2) be converted to x-2y-4 = 0?

Wrong
Double two on both sides
2y+2=-x+2
So x + 2Y = 0
Or y + 1 = 1 / 2 (X-2)
Double two on both sides
2y+2=x-2
So x-2y-4 = 0

The general implicit solution of the differential equation (x ^ 2 + y ^ 2) DX + 2xydy = 0 is?
A,(x^3)/3+x(y^2)=c
B,x^3+x(y^2)=c
C,(x^2)/2+x(y^2)=c
D,x^3-x(y^2)=c
Please explain the reason. How to solve a differential equation like this?

Let P (x, y) = x ^ 2 + y ^ 2, q (x, y) = 2XY, then α P / α y = α Q / α x, so the differential equation is a total differential equation
（x^2+y^2)dx+2xydy=0
x^2dx＋(y^2dx+2xydy)＝0
d(x^3/3)＋d(xy^2)＝0
d(x^3/3+xy^2)＝0
x^3/3+xy^2＝C
A.(x^3)/3+x(y^2)=c

The third quadrant P (m, - 1) is inside the circle x ^ 2-2x + y ^ 2-4 = 0, so find the value range of M

The third quadrant is m

To prove: (1) K / (K + 1)! = 1 / K! - 1 / (K + 1)! (2) 1 / 2! + 2 / 3! + n/(n+1)!=1-1/(n+1)!

1. K / (K + 1)! = 1 / K! - 1 / (K + 1)! Prove: K / (K + 1)! = ((K + 1) - 1) / (K + 1)! = 1 / K! - 1 / (K + 1)! 2. N / (n + 1)! = 1 / N! - 1 / (n + 1)! (n-1) / N! = 1 / (n-1)! - 1 / N! 2 / 3! = 1 / 2! - 1 / 3! 1 / 2! = 1 / 1! - 1 / 2! Add the above formulas and eliminate the same terms on the right side of the formula to get 1 / 2! + 2 / 3! + n/(n+...

If α is the angle of the third quadrant, cos α = 2m ^ 2-3 / 3-m, find the value range of M

Because α is the angle of the third quadrant, so - 1 ＜ cos α ＜ 0, so - 1 ＜ (2m ^ 2-3) / (3-m) ＜ 0 solution left: - 1 ＜ (2m ^ 2-3) / (3-m) (2m ^ 2-m) / (3-m) ＞ 0m (2m-1) (M-3) ＜ 0, so m ＜ 0 or 1 / 2 ＜ m ＜ 3 solution right: (2m ^ 2-3) / (3-m) ＜ 0 (2m ^ 2-3) (M-3) ＞ 0 (M + √ 6 / 2) (M - √ 6 / 2) (M -...)

Proving Π 3 ^ k / (3 ^ k - 1)

Let me have a try. Use the number reduction method
Let's prove that Π 3 ^ k / (3 ^ k - 1)

If the symmetric point m ′ of point m (2m + 1,3-m) about y axis is in the second quadrant, then the value range of M is___ ．

The symmetric point m ′ of ∵ point m (2m + 1, 3-m) about y axis is in the second quadrant, the ∵ point m (2m + 1, 3-m) must be in the first quadrant, ∵ 2m + 1 > 03-m > 0, the solution is: - 12 < m < 3

Find the limit (1) LIM (n - > ∞) ∫ (0,1) x ^ n / (1 + x) DX (2) LIM (n - > ∞) ∫ (n + K, n) SiNx / xdx (k > 0)

0 < xⁿ/(1 + x) < xⁿ
0 < ∫(0→1) xⁿ/(1 + x) dx < ∫(0→1) xⁿ dx = xⁿ⁺¹/(n + 1) |(0→1) = 1/(n + 1)
∵lim(n→∞) 1/(n + 1) = 0
∴lim(n→∞) ∫(0→1) xⁿ/(1 + x) dx = 0
0 ≤ |∫(n→n + k) (sinx)/x dx| ≤ ∫(n→n + k) |sinx|/|x| dx ≤ ∫(n→n + k) 1/n dx = k/n
∵lim(n→∞) k/n = 0
∴lim(n→∞) ∫(n→n + k) (sinx)/x dx = 0

The user inputs a positive integer n. if n is odd, the program calculates the sum of sequence 1 + 3 + 5 +... + n; if n is even, the program calculates 2 + 4 + 6 + +Sum of n

C language
main(){
int n,sum=0;
Printf ("/ N please enter a positive integer:);
scanf("%d",&n);
Printf ("the sum of the first n items of a sequence is / N")
if(n%2!=0){
for(int i = 2;i