1、1y ()nyy ()f x(,)f x y(,)f y y 2()ny(1)ny(2)ny 1()df xxC dx ()df xx 1C x()f x 1C dx 2C 2C()f xdx 3求求解解y 212xe 214xe y sin x 21C x cos x 12C x y 2xe()dx 2cosxex y 211sin22xexC()dx 12C 2C sin x 12C 2C 2C x 3C 218xecos.x 4F 22ddxmttFoT0FF 0(1)tFTddxt0(1)tFT tF()F t 0FFddxt 0Fm0,FtT 1 x0,0t 0,0t 2()2ttT
2、 1C 510,C 得得ddxt x 20,C 得得ddxt 201()2FtCTtm20()2FttmT 0Fm2(2t2C 3)6tT ddxt0 0t x0 0t x 0Fm2(2t3)6tT 6y y y p p (,)f x y(),p x(,)f x p(),p x1(,)x C y 1(,)x C y 1(,)x C dx 2C 72(1)x yy 2(1)x dpp ln p21(1)pCx即即13,C 得得y 323yxxC01,xy 21C 得得331yxx2x 1,3 y 2xp 2ln(1)x0 xy 3,03010 y y 0 x 0 x y (),p xddpx21
3、x 2dx x1ln,C 323x dpdx8y y ddyp p p y (,f yddyx(,)f y p 2xC(),p yddpxddpypddpy1(,),y C ddyx1(,),y C 1(,)y C dydx )y 9212p212ye 1C y dp p 10,C 得得ddyx ye 00,xy 再再由由21C 得得1yex 00,xy 01,xy 2dyeyddppy20,ye,yep e dyy 0 1 0y 0 2 ye yy 0,1 0 x 0 x(),p yy pddpydx,yex 2,C 010yy ydppln p 即即d yy ln y 0.2p dyy l
4、n y1d,Cx1C xy (),p ypddpy0,y ddyp p ddyx ddpxddpy1ln,C p 1,C yddyx1,C y 2lnC y 12C xC e2y 11dd x,(,)y yx (,F x,)y y y(,F x,)y y y0 dd x,(,)y yx 0,(,)y yx C yy 0.2y 21y2yyy 2y 0 yy d()d x0 yy 1C ln y()1C ln y1C x 2lnC y 12C xC e12()y x()yx()yy x 1,S2,S()y x12S1S 2,2yy 2S()yy x 设设曲曲线线(0)1,y 2S Pxy1S1o
5、yx(0)x 22tany 0()dxy tt 1221,SS2yy 0()dxy tt 1 2(),yyy(0)1,y 2()y 22()y(0)y 0,y2y2y 2()y ()yy x 0,2S 1,2()y y y y?113(),yp y 令令y dpp1,pC y 解解得得11,C ,yy 再再解解2,xyC e 21,C xye ddpypy2(),yyy(0)1,y 1(0)y 2S Pxy1S1oyxd,dppy2p dyy 14m 0,tyl 00ty d,dyvt 22ddytddvy ddvvy yoRlyddvt ddyt ddvmvy2km My dv v 212k
6、MvCy2d,kMyy 2112(),vkMyl2kMlyvly 即即d,dyvt dtd2lyyk Mly 注意注意“”号号 22ddyt 2y2ykm M15dt d2lyyk Mly 两端积分两端积分 t(0)ylzly2cos令令2lk M 21(sin2)22llzzCk M dyyly 2lk M 22cosdlzz 2(arccos)ylyyll 2C 2lk M 由于由于 y=R 时时 ,yg 由原方程可得由原方程可得 2g RkM 因此落到地面因此落到地面(y=R)时时y Rt y Rv 0,tyl 20,C 得得由由 2(arccos)2lytlyylk Ml的速度和所需时
7、间分别为的速度和所需时间分别为 2()g R lRl 21(arccos)2lRlRRlRgl16oxdx21,4Fd x g ,Fma 214d x g 2214kd gm 222d0dxk xt 0(0),xx(0)0 x 22d,dxmt 令令 ,xv 则则 ddvxt ddvx ddvvx 代入代入 2d,dvvk xx 分离变量分离变量 2dd,vvk xx 两端积分两端积分 212v 2221,vCk xddxt 222xk 1,2C 172221,vCk x0(0),xx(0)(0)0,vx 2210Ck x 22220(),vkxx220 xkxx 220d xxx 0arcs
8、inxx0(0),xx 2,2C 0sin()2xxk t 0cos.xk t d,kt 2,Ck t2,Tk ,k 2214kd gm 2214md g 214d g 9.8163.14 0.195(T)18oyx(0,1)A(1,0)(,)B x yvt,y 2v211d,xsyx ddt 21y 22ddyxx1vtyx 1xyyvt 211dxyx ddtvx y y ddxtddst ddtx21d1()2dyvx 22ddyxx21102y 1922ddyxx10,xy 11xy oyx(0,1)A(1,0)(,)B x yvt21102y 20内容小结内容小结 ()1.ny2.y
9、 (),yp x ddpyx 3.y (),yp y ddpypy (,),pf x p ddppy()f x(,)f x y(,)f y y(,),f y p2122gs cosF tan Moyx(a gs y 21y 1a F0FAsg1sa 0 x dx0,F,F 0F sinF y 1a 21y 0gsF 0)Fg 23y ,aOA 设0,xya 0 0 xy (),yp x 令令,ddxpy 则2d1pp 1dxa 2arshln(1)ppp 1arsh,xpCa0 0 xy 由由10,C 得得shxya 2ch,xyaCa0,xya 由由20C 得得chxyaa()2xxaaaee MoyxTHAsg211ya 可编辑感谢下感谢下载载
