Re: [微積] 一題微積分求解

[ejialan (eji)]

※ 引述《kattte (誠實面對自己吧!)》之銘言：
: 題目  y'-y*tan(x)=sin(2x)
: 令y=u*cos(x) ,y'=u'cos(x)-u*sin(x)
: u'cos(x)-u*sin(x)-u*cos(x)*tan(x)=sin(2x)
: u'cos(x)-u*sin(x)-u*sin(x)=2sin(x) cos(x)
:  u'=4sin(x) -> du=4sin(x)dx -> u=-4cos(x)
:  y=-4 cos^2(x)
: 請問...可以嗎?

令y = yh+yp

yh滿足yh'-yh*tan(x) = 0

dyh/dx - yh*tan(x) = 0

dyh/yh = tan(x)*dx

兩邊積分 => ln(yh) = -ln(cos(x)) + ln C = ln(C/cos(x))

取exp => yh = C/cos(x)

yp滿足yp'-yp*tan(x) = sin(2x) ...(1)

令yp = u/cos(x), yp' = (u'*cos(x)+u*sin(x))/cos(x)^2

代入(1)得 u'/cos(x) = sin(2x) = 2*sin(x)*cos(x)

u' = 2*sin(x)*cos(x)^2

u = -2*cos(x)^3/3

yp = u/cos(x) = -2*cos(x)^2/3

y = yh+yp = C/cos(x) - 2*cos(x)^2/3

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