推 S1010630 : 感謝大大指導QQ 08/05 01:27
→ remember : 這樣說的話，y=3/2 也是解 08/06 23:33
→ remember : 代進原式，兩邊都是-3/4 08/06 23:34
※ 引述《S1010630 (ggod)》之銘言：
: https://i.imgur.com/4l3YkuK.jpg
: 請問
: (a)小題要用何種解法，求各位提點
: （b)小題，小弟已經算出通解了；異解該如何得出
(a)
yy' = [y^2 + 2x^4 cos(x^2)]/x
u = y^2, v = x^2
=> du/dv - (1/v)u = 2v cos(v)
接下來就是一般作法
μ = 1/v
d(u/v)/dv = 2cos(v)
=> u = 2vsin(v) + cv
=> y^2 = 2x^2 sin(x^2) + cx^2
y(√π) = 0 => 0 = 2πsinπ + cπ => c = 0
=> y = +- x√[2sin(x^2)]
(b)
y' = y(2 - 3y)
=> [1/y - 1/(y - 2/3)]dy = 2dx
=> y/[3y - 2] = Aexp(2x) A > 0
異解用看的，y = 0顯然是正確的
當A = 0也可以得到y = 0
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