knrt.net
当前位置:首页 >> 用两边微分法求xy E∧x y >>

用两边微分法求xy E∧x y

xdy+ydx=e^(x+y)(dx+dy)

d(xy)=de^(x+y)xdy+ydx=e^(x+y)d(x+y)xdy+ydx=e^(x+y)(dx+dy)xdy+ydx=e^(x+y)dx+e^(x+y)dyxdy-e^(x+y)dy=e^(x+y)dx-ydx[x-e^(x+y)]dy=[e^(x+y)-y]dxdy=[e^(x+y)-y]dx/[x-e^(x+y)]

lnx=ylny 1/xdx=(lny+1)dy dy/dx=1/x(lny+1)

将y看成x的函数,等号两边同时对x求导.y + xy' = (y + xy')e^xy提出y':x(1 - e^xy)y' = y(e^xy - 1)因x,y均不为0(题目中等号右端指数函数恒大于0,故左端不为0),故(1 - e^xy)不为0,约去.y' = y/x

xy'-y=e^(x-1/x) 方程两边同时除以x^2得: y'/x-y/x^2=(1/x^2)*e^(x-1/x) 即 (y/x)'=(1/x^2)*e^(x-1/x) 积分得: y/x=∫(1/x^2)*e^(x-1/x)dx=q(x)+a 则得: y=x[q(x)+a] 关键是下面这个积分: q(x)=∫(1/x^2)*e^(x-1/x)dx

xy=e^(x+y) 两边对x求导 y+x(dy/dx)=e^(x+y)(1+dy/dx) 解得 dy/dx=[y-e^(x+y)]/[e^(x+y)-x]

∵1+sin(x+y)=e^(-xy),∴[1+sin(x+y)]e^(xy)=1,∴[1+sin(x+y)]′e^(xy)+[1+sin(x+y)][e^(xy)]′=0,∴[0+(x+y)′cos(x+y)]e^(xy)+[1+sin(x+y)][e^(xy)](xy)′=0,∴(1+y′)cos(x+y)+[1+sin(x+y)](x′y+xy′)=0,∴(1+y′)cos(x+y)+[1+sin(x+y)](y+xy′)=0,∴cos(x+y)+y′cos(x+y)

y=e^(-x/y)是这个吗?如果是则两边对x求导得y'=e^(-x/y)*(-x/y)'y'=e^(-x/y)*(-y+xy')/y^2整理即可

y+xy'=e^(x+y)(1+y') y+xy'=e^(x+y)+y'e^(x+y) y+xy'=xy+xyy' 再求导 y'+y'+xy''=y+xy'+yy'+x(y'^2+yy'') y+(x+y-2)y'+xy'^2+yy''=0

x*dy/dx+y=0的通解为y=C/x 用常数变易法,令原方程通解为y=C(x)/x 代入原方程,化简后可得C'(x)=e^x 积分得到C(x)=e^x+C 代回后即得原方程通解y=(e^x+C)x

网站首页 | 网站地图
All rights reserved Powered by www.knrt.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com