t + x dt/dx = t? - t
x dt/dx = t? - 2t
dt / (t? -2t) = dx /x 利用 1/ (t? -2t) = 1/2 [1 / (t -2) - 1 / t]
dt / [1 / (t -2) - 1 / t] = 2 dx /x
Ln((t-2) / t) = Lnx? + LnC
(t-2) / t = Cx ?
t = -2 / (Cx? -1)
即 y / x = -2 / (Cx? -1)
y = -2x / (Cx? -1)
代入條件當x = 1時， y(1) = 1，得到C = -1
特解為：y = 2x / (x? +1)
第三道：求二重積分∫∫√x^2+y^2 dxdy，其中積分區域D={(x,y)|x^2+y^2≤2x，0≤y≤x}。
參考答案：10/9√2
D={(x,y)|x?+y?≤2x，0≤y≤x}
===> D={(x,y)|(x-1)? + y?≤1，0≤y≤x}
===>方法1：先x後y：D={(x,y)|(x = y →1+√(1 - y?)，0≤y≤1}
方法2：極坐標：r? = 2rcosθ；即： θ = 0→π/4；r = 0→2cosθ
原積分 = ∫∫√x^2+y^2 dxdy
=∫{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)