推 monkps:原PO第二題看錯 是y-3 不過也讓我多學一題 謝謝 03/24 23:53
→ ytyty:沒看錯喔~移項過來變負號~所以變3-y 03/24 23:57
推 monkps:sorry 我看錯 鬼打牆 謝謝 03/25 07:35
※ 引述《monkps ()》之銘言：
: 1. xyy'=2y^2+4x^2 y(2)=4
y'=2x^(-1)y+4xy^(-1)
y'-2x^(-1)y=4xy^(-1)
Bernoulli n=-1
令u=y^[1-(-1)]=y^2
du dy
─ = 2y ─ = 2y[4xy^(-1)+2x^(-1)y] = 8x+4x^(-1)y^2
dx dx
du
─ -4x^(-1)u = 8x ...(1)
dx
∫-4x^(-1) dx
I.F.= e = x^(-4)
(1)乘以x^(-4)得
du
x^(-4) ─ -4x^(-5)u = 8x^(-3)
dx
d
─[x^(-4)u]= 8x^(-3)
dx
x^(-4)u =∫8x^(-3) dx =-4x^(-2) + c
x^(-4)*y^2 = -4x^(-2) + c
y(2)=4代入得
2^(-4)*4^2 = -4*[2^(-2)] + c
1 = -1 + c
c = 2
所求為x^(-4)*y^2 = -4x^(-2) + 2
: y-3
: 2.y'=------------
: x+y-1
(3-y)dx + (x+y-1)dy = 0...(1)
M=3-y，N=x+y-1
@M @N
─ = -1 ─ = 1
@y @x
@M @N
─ ≠ ─
@y @x
@M @N -2 -2
[ ─ - ─ ] /M = ─── = -g(y) g(y) = ──
@y @x 3-y y-3
-2
∫── dy 1
I.F.=e y-3 = ────
(y-3)^2
(1)乘以(y-3)^(-2)得
-(y-3)^(-1)dx + (y-3)^(-2)*(x+y-1)dy = 0必為exact
Ma=-(y-3)^(-1)，Na=(y-3)^(-2)*(x+y-1)
存在ψ(x，y)使得
@ψ
─ = Ma
@x
ψ= ∫Ma dx = ∫-(y-3)^(-1) dx =-(y-3)^(-1)*x+h(y)
@ψ
─ = (y-3)^(-2)*x+h'(y) = Na = (y-3)^(-2)*(x+y-1)
@y
= (y-3)^(-2)*x + (y-3)^(-2)*(y-1)
y-1 1 2
h'(y)= ──── = ─── + ───
(y-3)^2 y-3 (y-3)^2
1 2
h(y)= ∫─── + ─── dy = ln│y-3│- 2(y-3)^(-1) + k
y-3 (y-3)^2
ψ(x，y)=-(y-3)^(-1)*x+ln│y-3│- 2(y-3)^(-1) + k
=(-x-2)*(y-3)^(-1)+ln│y-3│+ k
所求 (-x-2)*(y-3)^(-1)+ln│y-3│= c
-x-2
即 ── + ln│y-3│= c
y-3
: 3. (t^2-16)y'+ty=2t
t 2t
y'+───y = ─── ...(1)
t^2-16 t^2-16
t
∫─── dt
t^2-16 (1/2)ln│t^2-16│
I.F.= e = e = (t^2-16)^(1/2)
(1)乘以(t^2-16)^(1/2)得
(t^2-16)^(1/2)y'+t*(t^2-16)^(-1/2)y=2t*(t^2-16)^(-1/2)
d
──[(t^2-16)^(1/2)y]=2t*(t^2-16)^(-1/2)
dt
(t^2-16)^(1/2)y = ∫2t*(t^2-16)^(-1/2)
(t^2-16)^(1/2)y = 2*(t^2-16)^(1/2) + c
y = 2 + c*(t^2-16)^(-1/2)
: 有勞各位了
: 謝謝
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