t + x dt/dx = t？ Equivalent to -ED
x dt/dx = t？ - 2t
dt / (t？ -2t) = dx /x usage 1/ (t? -2t)= 1/2[ 1/(t-2)- 1/t]
dt/[ 1/(t-2)- 1/t]= 2dx/x
Ln((t-2) / t) = Lnx？ + LnC
(t-2) / t = Cx？
t = -2 / (Cx？ - 1)
That is, y/x = -2/(Cx? - 1)
y = -2x / (Cx？ - 1)
When the substitution condition is x = 1, y( 1) = 1, C =-1 is obtained.
The special solution is: y = 2x/(x? + 1)
The third way: find the double integral ∫∫√ x 2+y 2dxddy, where the integral area d = {(x, y) | x 2+y 2 ≤ 2x, 0≤y≤x}.
Reference answer: 10/9√2
D={(x，y)|x？ +y？ ≤2x，0≤y≤x}
= = = & gtD={(x，y)|(x- 1)？ + y？ ≤ 1，0≤y≤x}
= = => method 1: x before Y: D = {(X, Y) | (X = Y → 1+√ (1-Y? )，0≤y≤ 1}
Method 2: polar coordinates: r? = 2COS θ, that is, θ = 0→π/4; r = 0→2cosθ
Original integral = ∫∫√ x 2+y 2dxddy
=∫{θ = 0→π/4}∫{r = 0→2cosθ} √r？ r drdθ
=∫{θ = 0→π/4}∫{r = 0→2cosθ} r？ drdθ
=∫{θ = 0→π/4} 8cos？ θ /3 dθ
=8/3∫{θ = 0→π/4} 1 - sin？ θ dsinθ
=8/3 [ sinθ - sin？ θ/3 ] {sinθ = 0→ 1/√2}
=8/3 [ 1/√2 - 1/(6√2)]
=8/3 [5/(6√2)]
= 10/(9√2)