Re: [微積] 一題微積分求解
※ 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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11/21 17:46, , 1F
11/21 17:46, 1F
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