A polar coordinate transformation problem R = sin θ, what does its rectangular coordinate equation look like? In definite integral. This is the upper limit of integral

A polar coordinate transformation problem R = sin θ, what does its rectangular coordinate equation look like? In definite integral. This is the upper limit of integral

Is r in R = sin θ polar? From R = sin θ, R * r = R * sin θ is x ^ 2 + y ^ 2 = y,
That is, x ^ 2 + (Y-1 / 2) ^ 2 = 1 / 4, so its rectangular coordinate equation is x ^ 2 + (Y-1 / 2) ^ 2 = 1 / 4

As shown in the figure, △ abd, △ ace are equilateral triangles

It is proved that: ∵ abd, △ ace are equilateral triangles, ∵ AC = AE, ad = ab. ∵ ∵ EAC = ∠ DAB = 60 °, ∵ EAC + ∠ BAC = ∠ DAB + ∠ BAC, i.e. ≌ EAB = ∠ CAD. In △ EAB and △ CAD, AE = AC, ≌ EAB = ∠ CAD, ab = ad, ≌ EAB ≌ CAD. ≌ be = CD

Triangle abd and triangle AEC are equilateral triangles. Prove be = DC

Let me draw first
Because it's an equilateral triangle
So Da = Ba, AE = AC,
Angle DAB = angle EAC = 60 degrees
Angle DAB + angle EAC = angle EAC + angle eac
So angle DAC = angle BAE
So triangle Abe is equal to triangle DAC (SAS)
So be = DC

ydy-ydx+xdy=0

ydy/dx-y+xdy/dx=0
(y+x)dy/dx=y
dy/dx=y/(y+x)=1/(1+x/y).1
Let t = x / y
x=yt
dx=tdy+ydt
Substituting 1
(1+t)dy=tdy+ydt
dy=ydt
1/ydy=dt
Simultaneous integral
lny+c=t
lny-x/y+c=0

In the following expression, a YDX + XDY B YDX XDY C xdx + YDY D xdx YDY is definitely not the total differential of a binary function

The answer should be B
Example of answer a: F (x, y) = XY
Example of answer C: F (x, y) = 0.5 x ^ 2 + 0.5 y ^ 2
Example of answer C: F (x, y) = 0.5 x ^ 2 - 0.5 y ^ 2

General solution of differential equation XDY YDX = y ^ 2 e ^ y dy

-(ydx-xdy)/y^2=e^ydy
d(x/y)+e^ydy=0
therefore
x/y+e^y=C

Solving homogeneous differential equation: (x ^ 2 + y ^ 2) DX = xydy

Let u = Y / x, then y = Xu, dy / DX = u + + X * Du / DX
U + X * Du / DX = u + 1 / u. so UDU = DX / X. integral on both sides 1 / 2 * u ^ 2 = LNX + lnc. Substituting u = Y / x, the general solution y ^ 2 = 2x ^ 2ln (Cx) is obtained
In addition, X ≡ 0 is also the solution of the differential equation

General solution of differential equation xydy + (y ^ 2 + 1) DX = 0

If the equation is YDY / (Y & # 178; + 1) = - DX / x with separable variables, the integral on both sides is: ∫ Y / (Y & # 178; + 1) dy = - ∫ 1 / X DX is: (1 / 2) ∫ 1 / (Y & # 178; + 1) d (Y & # 178;) = - ln | x | ln (Y & # 178; + 1) = - 2ln | x | + LNC, so: Y & # 178; + 1 = C / X & # 178

How to find the general solution of the differential equation (x ^ 2 + 2XY) DX + xydy = 0?
Detailed steps to solve the problem, my little brother, thank you very much!

Dy / DX = - (x ^ 2 + 2XY) / (XY) = - (x + 2Y) / y = - X / Y-2 let u = Y / x, then y = Xu, y '= u + Xu' is substituted into the original equation: U + Xu '= - 1 / u-2xu' = - 1 / u-2-u = - (U + 1) ^ 2 / uudu / (U + 1) ^ 2 = - DX / XDU * [1 / (U + 1) - 1 / (U + 1) ^ 2] = - DX / X integral: ln | U + 1 | + 1 / (U + 1) = - ln | x | + C1 ln | Y / x + 1 | + 1 / (Y / X / X / Y / x) = - DX

The general solution of the differential equation (Y-1) DX xydy = 0,

(y-1)dx-xydy=0
dx/x=ydy/(1-y)
That is DX / x = [1 + 1 / (Y-1)] dy
The integral on both sides is ∫ DX / x = ∫ [1 + 1 / (Y-1)] dy
lnx=y+ln(y-1)+C1
x=C(y-1)e^y