求(2y^2-9xy)dx+(3xy-6x^2)dy=0之通解，已知I(x,y)=x^ay^b

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)

(6xy^2-y^3)dx+(6x^y-3xy^2)dy=d(3x^y^-xy^3),∴原式=（3x^y^-xy^3)|,=(9x^-7x)|=9*7-7=56.再问：原式==（3x^y^-xy^3)|,3x^y^是什么意思x^y^??再答

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

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=3∫xdxlny-ln(y+3)=1.

别人一般问一道题,你一下子5道?我给你个提示：1.所有5道题全部可以化成y'=f(y/x)的形式.比如5：:y’=√(1-y^2/x^2)+y/x2.设y/x=uy=xuy'=u+xu',代入：u+xu'=f(u)比如5：:u+xu'=√(

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。

原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x²y²+12xy&su

dy/dx=(xy+3x-y-3)/(xy-2x+4y-8)=(x-1)(y+3)/(x+4)(y-2)再问：然后呢？再答：(y-2)dy/(y+3)=(x-1)dx/(x+4)已经是变量分离方程，两边积分可求解再答：(y+3-5)dy/(

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=-cosxy（ydx+xdy）+2ydy=0

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

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²)-x]

dy/dx=x(y²+1)/(y-xy²)

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)

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.(1)∵方程(1)一阶线性微分方程∴由

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/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^3d(1/(3x^3))+d(1/(2x^

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.(1)∵方程(1)一阶线性微分方程∴由

xy/x+y=6∴xy=6(x+y)3x-2xy+3y/-x+3xy-y=[3x-12(x+y)+3y]/[-x+18(x+y)-y]=-9(x+y)/17(x+y)=-9/17再问：你在啊？再答：在再问：再问你一道题嘞。。。行么？再答：可

参考一下

y=(-x^2+Cx)^(1/3),C为任意常数解题步骤：3xy^2dy=(y^3-x^2)dx,(3xy^2)*y'=y^3-x^2,又[(y^3)/x]'=[(3xy^2)*y'-(y^3)]/(x^2)=-1,知(y^3)/x=-x+