搜狗做题网dx=4,dy=1,xy=0.6,d(x-y)=
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观察知,y=x是方程的特解为求通解,令y=x+t,代入原方程得(1+x^2)(1+t')dx=(1+x^2+xt)dx化简得dt/t=xdx/(1+x^2)所以,t=C(1+x^2)^(1/2)所以,
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
dy=xydx1/ydy=xdxln|y|=x²/2+C∴dy/dx=xy的通解为y=±e^(x²/2+C)e^(x²/2+C)表示±e的(x²/2+C)次方再
dy/dx=e^(xy)dy/e^y=e^xdx两边积分得-e^(-y)=e^x+C再问：你这样右边是e^(x+y)啊再答：噢令xy=p两边求导得y+xy'=p'y'=(p'-y)/x=(p'-p/x
两端对x求导得y+xy'=e^(xy)*(y+xy')整理即可得dy/dx=y再问：y'=y+e^xy/e^xy-x?再答：是的啊，就是这样啦。
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
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)]
两边对x求导得cosx+y'cosy=y+xy'解出来y'就可以了再问：z=f(xy^2,x^2y)求δz/δx,δz/δy这个呢再答：令u=xy^2,v=x^2yδz/δx=f'u*u'x+f'v*
就是把这dydx转为求导前的式子,然后再求导一遍验证一下对错.再问：就是算到最后有个积分搞不出来。求过程。
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)
x/[sec(xy)-y]dx/dy.
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
分离,有dy/(2-y)=2xdx,d(2-y)=-dy,所以-d(2-y)/(2-y)=2xdx,两边积分,有-ln|2-y|=x^2+C>=0,所以ln|2-y|=0,y=1或3,x=0,C=0
∵(1-x^2)dy/dx+xy=1==>(1-x^2)dy+xydx=dx==>dy/(1-x^2)^(1/2)+xydx/(1-x^2)^(3/2)=dx/(1-x^2)^(3/2)(等式两端同除
y=-1/4-(1/4)x+(3/8)tan(6x-c)再问：求过程
dx/dy+xy=-1积分因子：exp(∫ydy)=exp(y²/2)=e^(y²/2)dx/dy•e^(y²/2)+xy•e^(y²/
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
答：xy=x-e^(xy)e^(xy)=x-xy=x(1-y)两边对x求导：(xy)'e^(xy)=1-y-xy'(y+xy')e^(xy)=1-y-xy'ye^(xy)+xy'e^(xy)+xy'=