The general solution of (1 + x ^ 2) dy + 2xydx = cotxdx

The general solution of (1 + x ^ 2) dy + 2xydx = cotxdx

(1+x²)dy+(2xy-cotx)dx=0
∂(1+x²)/∂x=∂(2xy-cotx)/∂y=2x
This is a total differential equation
∫(1+x²)dy=y+x²y+Φ(x)
∫(2xy-cotx)dx=x²y-ln|sinx|+Ψ(y)
u(x,y)=y+x²y+Φ(x)=x²y-ln|sinx|+Ψ(y)
The comparison coefficient is Φ (x) = - ln | SiNx |, ψ (y) = y
u(x,y)=y+x²y-ln|sinx|
So the general solution of the original equation is y + X & # 178; y-ln | SiNx | = C

In the rectangular coordinate system, the set of points in the second quadrant and the fourth quadrant is expressed as?
tt

{(x, y) | x, y different sign}

In the rectangular coordinate system xoy, if the vertex of angle α is at the origin of the coordinate system, the starting edge is the positive half axis of the x-axis, and the ending edge passes through the point (1, 2),
In the rectangular coordinate system xoy, if the vertex of angle α is at the origin of the coordinate system, the starting edge is the positive half axis of X axis, and the ending edge passes through the point (1,2 √ 2). If points a and B are the moving points on the starting edge and ending edge of angle α respectively, and ab = √ 3, the maximum value of △ ABO area can be obtained
come on..

Let OA = m, OB = n, then | ab | & sup2; = 3 = M & sup2; + n & sup2; - 2mncos α = M & sup2; + n & sup2; - (2 / 3) Mn, then 3 + (2 / 3) Mn = M & sup2; + n & sup2; ≥ 2Mn, then Mn ≤ 9 / 4, so area s = (1 / 2) mnsin α ≤ 3 √ 2 / 4

In space rectangular coordinate system, given point a (1,2,4), point B and point a are symmetric about y axis, and point C and point a are symmetric about plane xoz, the distance between point B and point C is calculated

In the space rectangular coordinate system, the symmetric point of point a (1,2,4) about the plane xoz is C (1, - 2,4), and the symmetric point of point a (1,2,4) about the X axis is B (- 1,2, - 4), then the distance between B and C is: (1 + 1) 2 + (2 + 2) 2 + (4 + 4) 2 = 221

Symmetrical point in space rectangular coordinate system
The point m (x, y, z) is a point in the space rectangular coordinate system oxyz. Write the coordinates of the store satisfying the following conditions
(1) On plane xoy symmetry with point m
(2) On plane YOZ symmetry with point m
(3) On plane xoz symmetry with point m

1.x,y,-z
2.-x,y,z
3.x,-y,z

The symmetry of point and point in space rectangular coordinate system
The title is as follows:
What are the coordinates of the symmetric point m 'of point m (1, - 4,3) with respect to point P (4,0, - 3)?
PS: what are the principles?

If you multiply the coordinates of P by 2 and subtract the coordinates of M, you get the symmetry point, which is (7,4, - 9). Thank you

How to find the symmetric point of a point about a straight line in space rectangular coordinate system?

The slope of the line is K
Q(x,y)P(a,b)
((x + a) / 2, (y + b) / 2) satisfies the equation
[(b-y)/(a-x)]k=-1

How to find the symmetric point of one point about another in the space rectangular coordinate system
What are the coordinates of the symmetric point m 'of point m (1, - 4,3) with respect to point P (4,0, - 3)?
How should it be calculated? Is it (7,4, - 9)

The coordinate of point P multiplied by 2 minus the coordinate of point m is the coordinate of another point

In space rectangular coordinate system, the problem of finding a point symmetrically with respect to a straight line is solved,
If point m and n (2,5,0) are symmetric with respect to line L, the coordinates of point m are obtained
The equation is x-y-4z + 12 = 0 2x + y-2z + 3 = 0
If the calculation is troublesome, can you tell me the idea

Point m and n (2,5,0) are symmetric with respect to line L, and the coordinates of point m are obtained
The equation is x-y-4z + 12 = 0 2x + y-2z + 3 = 0
The symmetric point of a point with respect to a line
It can be seen that the midpoint of two points is on the symmetrical line, and the line determined by two points is perpendicular to the symmetrical line
The equation is x-y-4z + 12 = 0 2x + y-2z + 3 = 0
The direction vector is (1, - 1,0.5)
Let the coordinates of point m be (x, y, z)
Then the coordinates of the midpoint of M and n (x / 2 + 1, Y / 2 + 2.5, Z / 2)
The direction vector of the linear equation where Mn is located is (x / 2-1, Y / 2-2.5, Z / 2)
By substituting the midpoint coordinates into the linear equation, two vectors are perpendicular
The coordinates of m point can be calculated

In the space rectangular coordinate system, if the point P (1,0,1) Q (4,3, - 1) has a point m on the z-axis, let | MP | = | MQ |? If it exists, the coordinate of point m is obtained;

Let m (0,0, z)
There is MP ^ 2 = 1 + (Z-1) ^ 2
MQ^2=4^2+3^2+(z-1)^2=5^2+(z+1)^2
So (Z + 1) ^ 2 + 25 = 1 + (Z-1) ^ 2
z=-6
M(0,0,-6)