推 topee:利害 04/07 21:59
※ 引述《wellilin (QQ12)》之銘言：
: 1 y'tanx-2y=4 y(π/2)=1
: ans: y= -2+3sinx^2
dy - 2 /tanx ydx = 4 / tanx dx
I = exp (∫-2cosx/sinx dx) = exp (-2lnsinx) = sin^-2x
dysin^-2x = ∫4 *sin^-2x *cosx/sinx dx
dysin^-2x = ∫4 / sin^-3x dsinx
dysin^-2x = -2sin^-2x + c
y = -2 + c sin^2x
y(π/2)=1 1 = -2 + c c = 3
y= -2 + 3 sinx^2 #
: 2. xdy-(y+xy^3(1+lnx))dx=0
: ans: 1/y^2=-4/9x-2/3lnx+c/x^2
: 3.
: (xy^3-y^3-x^2e^x)dx+3xy^2dy=0
: ans: y^3=x/2e^x+cxe^-x
3. xy^3dx -y^3dx+xdy^3 = x^2 e^xdx
xy^3dx + x^2 d y^3/x = x^2 e^xdx
除x^2
d(y^3/x) + (y^3/x) dx = e^x dx
I = exp [∫1dx] = e^x
d(y^3/x)e^x = e^2x dx
(y^3/x)e^x = 1/2 e^2x + c
y^3 = x/2 e^x + cxe^-x #
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※ 編輯: funtsung 來自: 114.26.177.86 (04/07 22:00)