[理工] [工數]-ODE
1.
2
y = 2xy' + y(y')
答案是
y = +- ix 奇異解
2
y = +- 根號 2c1*x + c1
我的做法
+ 2 2
-2x - 根號4x + 4y
y' = ---------------------
2y
+
-2x - (2x+2y)
y' = --------------
2y
-2x+2x+2y -2x-2x-2y -2x-y
y' = ---------- = 1 or y' = ------------ = ------
2y 2y y
^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^
dy=dx ydy = -2xdx - ydx
y = x+c 然後就卡了...
接下來有請高手了
2.
1 2
y = xy' + ---(y')
2
1 2
y = cx + ---c 是通解
2
1 2
y = - ---x 為奇異解
2
奇異解我算出來 但是通解怪怪的
p = y'
dp dp
p--- + p + x--- = p
dx dx
dp
(p+x)--- = 0
dx
p+x=0 算出來 是 奇異解 我省略
dp
--- = 0
dx
dp=0
p = c1
y' = c1
dy = c1dx
y = c1x + c2
y'=c1 帶入ode
1 2
---(c1) + xc1 -(c1x + c2) = 0
2
1 2
---c1 - c2 = 0
2
怎麼感覺根答案有點出入
我錯在哪@@
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