工數微分方程~各位大大幫幫我吧！

沒有I.C. or B. C.嗎?

2010-06-24 15:14:09 補充：
2y'*y"=2e^(2y)*y'
兩邊對x積分,then
y'^2= e^(2y)+A (A is an integral constant.)
∫ dy/√[e^2y)+A] = +/- ∫ dx

if A=0 then 1/e^y = +/- x +B (B is an integral constant.)

if A>0, A=p^2 (p>0) then ∫ e^(-y)/√[1+p^2 e^(-2y)] dy = +/- x +B
(-1/p) ln| exp(-y) + √[1/p^2+ exp(-2y)] | = +/- x +B
or p exp(-y)+√[1+ p^2 exp(-2y)] = B'*exp( +/- px)
or (-1/p) arcsinh[ p exp(-y)] = +/- x + B
or p exp(-y)= +/- sinh(px+B')

if A<0, A=-p^2, p>0, then ∫exp(-y)/√[1-p^2 exp(-2y)] dy = +/- x+ B
(-1/p)arcsin[p exp(-y)]= +/- x+ B
or p exp(-y)= +/- sin(px+B')