搜搜做题作业网(x y)dx-(x-y)dy x^2 y^2
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/22 14:23:40
解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²
x^2*dy/dx=xy-y^2dy/dx=y/x-y^2/x^2u=y/xy=xuy'=u+xu'代入：u+xu'=u+u^2xu'=u^2du/u^2=dx/x-1/u=lnx+lnCCx=e^(
方法一（微分法）d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即（xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²
两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)
结果当然可以写成:|(y-2x)^3=C(y-x)^2,C为待定常数,解曲线为下面是具体求解过程：
令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³
ydx/dy+x=(e^x)(e^y)dx/dy+(e^x)(e^y)dx/dy=[(e^x)(e^y)-x]/[y-(e^x)(e^y)]dx/dy=(xy-x)/(y-xy)dx/dy=x(y-1
答：dy/dx=1+x+y^2+xy^2y'=(1+x)(1+y^2)y'/(1+y^2)=1+x(arctany)'=1+x积分得：arctany=x+x²/2+Cy=tan(x+x
cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]
令y/x=u,dy=u+xdu,原方程化为:u+xdu/dx=x/(u^2)+u,即du/dx=1/(u^2)通解为:y=x*[(3x+3c)^(1/3)]
dy=xy*dx+x*dx+y*dx两边同时积分y=y*1/2*x*x+1/2*x*x+xy(1-1/2*x*x-x)y=1/2*x*xy=x*x/(2-2x-x*x)+C*是乘号x*x是x的平方
∵dy/dx=(x+y^3)/(xy^2)==>xy^2dy=(x+y^3)dx==>y^2dy/x^3=dx/x^3+y^3dx/x^4(等式两端同除x^4)==>d(y^3)/(3x^3)+y^3
令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³
显然不是啊
dx/dt=6tdy/dt=4costdy/dx=(dy/dt)*(dt/dx)=(4cost)*(1/6t)=2cost/3tdx/dy=3t/2cost再问：sin4t的导数是4cost吗？我们老
两边同时求导x+x(dy/dx)+1*(dy/dx)/y+1/x=0合并同类项dy/dx=-y/x
令y=xuy'=u+xu'代入方程：u+xu'=u^2/(u-1)xu'=u/(u-1)du(u-1)/u=dx/xdu(1-1/u)=dx/x积分;u-ln|u|=ln|x|+C1e^u/u=Cxe
你要求什么?
令e^(xy)=u,y=lnu/xDy/dx=[(x/u)*(du/dx)-lnu]/x²,∴(1/ux)*(du/dx)-lnu/x²+lnu/x²=u即du/u