Y = SiNx is the solution of the differential equation y "+ y = 0 Right and wrong

Y = SiNx is the solution of the differential equation y "+ y = 0 Right and wrong

correct
（sinx）’=cosx
（cosx）’=-sinx
therefore
（sinx）”=-sinx

General solution of equation y '' = (1 + y '* y') / 2Y

Let y ′ = P, then y ′′ = PDP / Dy, and substitute it into the original equation to get PDP / dy = (1 + P & sup2;) / (2Y). (1)
Solution (1) gives P = ± √ (c1y-1), (C1 is the integral constant)
‖ y ′ = = ± √ (c1y-1), (C1 is the integral constant)
dy/√（C1y-1）=±dx.（2）
The solution (2) is 2c1y = (C2 ± c1x) & sup2; + 1, (C1, C2 are integral constants)
So the general solution of the equation y '' = (1 + y '* y') / 2Y is:
2c1y = (C2 ± c1x) & sup2; + 1, (C1, C2 are integral constants)

Finding the general solution of the equation: (x ^ 2-1) y '+ 2XY cosx = 0)

It can be reduced to the first order non-homogeneous equation dy / DX + 2XY / (x ^ 2-1) = cosx / (x ^ 2-1), where p (x) = 2XY / (x ^ 2-1), q (x) = cosx / (x ^ 2-1), and the final result is y = (x ^ 2-1) (SiNx + C)

If the center of circle C and point (1,0) are symmetric with respect to the line x + y = 0, the line 3x + 4y-11 = 0 intersects with circle C at two points a and B, and / AB / = 6, the equation of circle is?
As detailed as possible. As soon as possible, urgent

Through (1,0) vertical x + y = 0 line L
50: Y = X-1, and X + y = 0 intersection P (1 / 2, - 1 / 2)
C（X,Y）
(x+1)/2=1/2,(y+0)/2=-1/2
Center C (0, - 1)
C to the straight line 3x + 4y-11 = 0 distance D:
D=|0-4-11|/5=3
（AB/2)^2+D^2=R^2
R^2=18
The equation of circle is as follows:
X^2+(Y+1)^2=18

When a circle passes through the point P (- 4,3), the center of the circle is only on the line 2x-y + 1 = 0 and the radius is 5, the equation of the circle is obtained

The center of the circle is on the line 2x-y + 1 = 0
Let the center of the circle be (x0,2x0 + 1)
Its distance from P (- 4,3) is 5,
That is, D & # 178; = (2x0-2) &# 178; + (x0 + 4) &# 178; = 5 & # 178;,
The solution is x0 = ± 1,
The center of the circle is (- 1, - 1) or (1,3),
Then the equation of circle is (x + 1) &# 178; + (y + 1) &# 178; = 25 or (x-1) &# 178; + (Y-3) &# 178; = 25

The polar coordinate equation mpcos & sup2; Q + 3psin & sup2; q-6cos q = 0 of the curve is transformed into rectangular coordinate equation, and the shape of the curve is explained

P = √ (x ^ 2 + y ^ 2) PCOSQ = xpsinq = ympcos & sup2; Q + 3psin & sup2; q-6cosq = 0m (PCOSQ) & sup2 / / P + 3 (psinq) & sup2 / / p-6pcosq / P = 0mx & sup2 / / P + 3Y & sup2 / / p-6x / P = 0. The rectangular coordinate equation is MX & sup2; + 3Y & sup2; - 6x = 0. When m is not 0, m (x-3 / M) & sup2; + 3Y & su

It is known that the polar coordinate equation of curve C is p = 8cosa / 1-cos2. The polar coordinate equation of curve C is transformed into rectangular coordinate equation
Let a (x 1, Y 1) B (x 2, y 2) find the value of Y 1y 2

ρ = 8cos θ / (1-cos2), (1-cos2) ρ & # 178; = 8 ρ cos θ, i.e. (1-cos2) (X & # 178; + Y & # 178;) = 8x. Is the title right?

It is known that hyperbola C passes through P1 (2,3), P2 (radical 2, radical 3)
1. Find the equation of hyperbola C
2, if P is a point on the left branch of the hyperbola, and ∠ f1pf2 = 60 °, calculate the area s of △ f1pf2

x^2/a^2-y^2/b^2=1
Bring in P1 and P2 to solve a ^ 2 and B ^ 2
Hyperbola C: x ^ 2 / 1-y ^ 2 / 3 = 1
Let P (x0, Y0)
Then s △ f1pf2 = 1 / 2 * | Y0 | * 4 = 2 | Y0|
P (x0, Y0) satisfies two equations
x0^2/1-y0^2/3=1……………………………………………………………………………… one
From ∠ f1pf2 = 60 ° we get | f1p | * | PF2 | * cos60 ° = vector f1p * vector PF2
That is, root ((x0-2) ^ 2 + Y0 ^ 2) * root ((x0-2) ^ 2 + Y0 ^ 2) * cos60 ° = (x0-2) (x0 + 2) + Y0 ^ 2 two
The simultaneous equations 1 and 2 can be solved to obtain | Y0 | = 1.5 times the root sign 3 (the other solution | Y0 | = 0.5 times the root sign 3 is rounded off, because ∠ f1pf2 = 60 degree)
It's an acute angle, so choose a larger one
So, s △ f1pf2 = 1 / 2 * | Y0 | * 4 = 2 | Y0 | = 3 times root 3
There is a formula for the area of hyperbolic focus triangle
If ∠ f1pf2 = θ, then s △ f1pf2 = B ^ 2 * cot (θ / 2) or s △ f1pf2 = B ^ 2 * / Tan (θ / 2)

Given the hyperbolic point P1 (- 2,3 radical 5 / 2), P2 (4 radical 7 / 3,4), the hyperbolic standard equation is solved

Let the hyperbolic standard equation be X & # 178 / / a-y & # 178 / / b = 1
Then 4 / a - (45 / 4) / b = 1, (112 / 9) / A-16 / b = 1
A = - 16, B = - 9
So the hyperbolic standard equation is Y & # 178 / 9-x & # 178 / 16 = 1

Find the standard equation of hyperbola passing through points (3, - 4, radical 2), (9 / 4,5),

Let the hyperbolic equation be MX ^ 2 + NY ^ 2 = 1,
Substituting it, we can get 9m + 32n = 1, 81m / 16 + 25N = 1,
The solution is m = - 1 / 9, n = 1 / 16,
Therefore, the hyperbolic equation is - x ^ 2 / 9 + y ^ 2 / 16 = 1