lim y^2cos(xy)dy

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^(

分离得到：dy/y=2xdx两边积分：ln|y|=x^2+C1y=±e^c1 *e^x^2  =Ce^x^2 (C =±e^c1) 图片如下

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

对方程取导数y+x(dy/dx)+(dy/dx)=0(dy/dx)(x+1)=-ydy/dx=(-y)/(x+1)

先求解dy/dx=2xy,得到：dy/y=2xdx,所以ln|y|=x^2+c,即y=Cexp(x^2),其中C为常数,此时再用常数变易法,设y=C(x)exp(x^2),代入原式可得C(x)=C0-

=-[ysin(xy)+2e^(2x+y)]/[ysin(xy)+e^(2x+y)]*(dx)再问：麻烦给我写出解的过程。。再答：等式两边取对数，得：d[e^(2x+y)]-d[cos(xy)]=0(

f(x,y)=e^(x+y)+cos(xy)=0      //: 利用隐函数存在定理：f 'x(x,y)=e^

先求dy/dx+2xy=0的解:dy/y=-2xdx,--->lny=-x^2+C=-ln(e^(x^2))+lnC=ln(C*e^(-x^2)),即y=C*e^(-x^2).然后令y=C(x)*e^

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

直接分离变量就可以了dx/x=2ydylnx=y^2+C即x=C1*e^(y^2)

y^2=(xy-x^2)dy/dxy^2/x^2=(y/x-1)dy/dxy/x=udy=udx+xduu^2=(u-1)(u-xdu/dx)u^2/(u-1)=u-xdu/dxxdu/dx=u-u^

cos(xy)=x两边对x求导：-sin(xy)[y+xy']=1y+xy'=-1/sin(xy)xy'=-y-(1/sin(xy))y'=[-y-(1/sin(xy))]/x

三种方法1式中同时对x求导-（y+xy‘）cosxy+2yy'=0解出y’2式中同时取微分d{sin(xy)+y^2-e^2}=dsin(xy)+dy^2-de^2=-cosxydxy+2ydy=-c

你好!两边对x求导：e^(xy)*(y+xy')-y^2=y'cosy解得y'=(y^2-ye^(xy))/(xe^(xy)-cosy)

dx/dy-3xy=xy^2dx/x=(y^2+3y)dy两边积分得：lnx=y^3/3+3y^2/2+c==>x=exp(y^3/3+3y^2/2+c)=Cexp(y^3/3+3y^2/2)C常数

x=0时,代入方程得：1+1=y,得：y=2对x求导：(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得：2=y'故dy(0)=2dx

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?

dy/dx=xy²+3xydy/dx=x(y²+3y)∫1/[y(y+3)]dy=∫xdx(1/3)∫(3+y-y)/[y(y+3)]dy=∫xdx∫[1/y-1/(y+3)]dy