Consider the equation Ux+Uy=U^(1/2)，derive the general solution
U(x,y)=(x+f(x-y))^2 / 4，observe that the trival solution U(x,y)=0 is not covered by the general solution.
一題微分方程的問題
2010-04-22 7:19 am
回答 (2)
2010-04-22 8:11 am
✔ 最佳答案
Let ξ=x+y, η=x-yU^(1/2)=∂U/∂x+∂U/∂y=∂U/∂ξ+∂U/∂η+∂U/∂ξ-∂U/∂η
then ∂U/∂ξ= 0.5U^0.5 -----(A)
Suppose U≠0, then
∫U/U^0.5 dU=∫ 0.5 dξ
2U^0.5= 0.5ξ+0.5f(η) (Note: 0.5f(η) is integral constant)
so U=[(x+y)+f(x-y)]^2/4
Note: U=0 is another solution of (A)
2010-04-22 12:07 pm
題目本身有矛盾. U(x,y)=(x+f(x-y))^2 / 4 is a solution,for any function f, to the equation 是一回事, 只要它有漏掉一個解它就不能叫通解. 事實上它還可以有其他形式的解(包括一任意單變數函數). 看天助的回答即瞭.
收錄日期: 2021-04-30 14:41:49
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