求微分方程的通解 y"-xy=0

发布网友 发布时间：2024-09-25 20:18

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热心网友 时间：2024-10-03 14:36

∵xy"+y'=0 ==>xdy'/dx+y'=0
==>dy'/y'=-dx/x
==>ln│y'│=-ln│x│+ln│C1│ (C1是积分常数)
==>y'=C1/x
∴y=∫C1/xdx
=C1ln│x│+C2 (C2是积分常数)
故原微分方程的通解是y=C1ln│x│+C2 (C1,C2是积分常数)。

xy'-y=0
xdy/dx=y
dy/y=dx/x
两边同时积分得
lny=lnx+lnc
y=cx.

热心网友 时间：2024-10-03 14:35

它属于微分方程幂级数的解法，它的通解一般都可设为y=x^a(a0+a1*x+a2*x^2+a3*x^3+........................)其中a,a0,a1,a2........为待定常数。其证明过程有点复杂，通常我们只需记着就行。推荐你看《高等数学微分方程解法》。上面有详细证明过程。

热心网友 时间：2024-10-03 14:39

该方程可以令y=e^t,可得 t"+(t')^2-x=0 ;再令t'=u,可转化为一阶微分方程 u'+u^2-x=0
该方程是Riccati方程,一般没有初等解法，只能用级数解法 如上面几位同学的解法

热心网友 时间：2024-10-03 14:31

该微分方程只能用级数解法

热心网友 时间：2024-10-03 14:38

Assume that y = Σ(n = 0 to ∞) a(n) x^n.

Then, y'' - xy
= Σ(n = 2 to ∞) n(n-1) a(n) x^(n-2) - x Σ(n = 0 to ∞) a(n) x^n
= Σ(n = 2 to ∞) n(n-1) a(n) x^(n-2) - Σ(n = 0 to ∞) a(n) x^(n+1)
= Σ(n = 0 to ∞) (n+3)(n+2) a(n+3) x^(n+1) - Σ(n = 0 to ∞) a(n) x^(n+1)
= 2a(2) + Σ(n = 0 to ∞) [(n+3)(n+2) a(n+3) - a(n)] x^(n+1) = 0

With a(0) and a(1) arbitrary, we have
(i) 2a(2) = 0 ==> a(2) = 0

(ii) For n ≥ 0, we have (n+3)(n+2) a(n+3) - a(n) = 0
==> a(n+3) = a(n) / [(n+3)(n+2)].

Since the recurrence is in steps of 3, we have three cases.
a(3) = a(0)/(3 * 2) = 1a(0)/3!
a(6) = a(3) / (6 * 5) = 1a(0) / [6 * 5 * 3!] = (1 * 4) a(0) / 6!
...
a(3k) = (1 * 4 * ... * (3k - 2)) a(0) / (3k)!
-------------
a(4) = a(1)/(4 * 3) = 2 a(1)/4!
a(7) = a(4) / (7 * 6) = 2 a(0) / [7 * 6 * 4!] = (2 * 5) a(1) / 7!
...
a(3k+1) = (2 * 5 * ... * (3k - 1)) a(1) / (3k+1)!
-------------
a(2) = 0 ==> a(5) = a(2)/(5 * 4) = 0
...
a(3k+2) = 0.

Therefore, the general solution is
y = a(0) Σ(k = 0 to ∞) (1 * 4 * ... * (3k - 2)) x^(3k)/(3k)!
+ a(1) Σ(k = 0 to ∞)(2 * 5 * ... * (3k - 1)) x^(3k+1) / (3k+1)!