(6xy^2-y^3)dx (6x^2y-3xy^2)dy

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(6xy^2-y^3)dx (6x^2y-3xy^2)dy
x^2+xy+y^3=1,求dy/dx

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

微分方程的两道题(dy/dx)-2xy=y+4x-2y''+2y(y')^3=0

1.(dy/dx)-2xy=y+4x-2dy/dx=(2x-1)(y+2),dy/(y+2)=(2x-1)dx,ln(y+2)=x^2-x+C,y=e^(x^2-x+C)-2.2.y''+2y(y')

dy/dx=(xy+3x-y-3)/(xy-2x+4y-8) 微分方程怎么求呀,求教,

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

x^2-xy-6y^2+3xy+y+2

3xy是3x原式=(x-3y)(x+2y)+3x+y+2十字相乘x-3y2×x+2y1所以原式=(x-3y+2)(x+2y+1)

求下列微分方程的解(1)(xy+x^3y)dy-(1+y^2)dx=0 (2)(y^2-6x)y'+2y=0

(1)(xy+x^3y)dy-(1+y^2)dx=0(xy+x^3y)dy=(1+y^2)dx分离变量整理得:y\(1+y^2)dy=1\x(1+x^2)dx整理:y\(1+y^2)dy=1\x-x\

解微分方程 (x^2y^3+xy)dy=dx

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

求由方程y^2-3xy+6=0所确定的隐函数的导数dy/dx

方程两边同时对x求导得2yy'-3(y+xy')=0整理化简得y'=3y/(2y-3x)即dy/dx=3y/(2y-3x)

∫ (6xy^2-y^3)dx+(6x^y-3xy^2)dy

(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)

dy/dx=(x^4+y^3)/xy^2

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

∵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

下面都是求微分方程的通解:1、(y^-2xy)dx+x^2dy=0 2、(x^2+y^2)dy/dx=2xy 3、xy’

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+x

微分方程求解 (x^2y^3+xy)dy=dx

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

dy/dx=3xy+xy^2.求y.

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。

已知xy/x+y=6,求3x-2xy+3y/-x+3xy-y的值.

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

一道简单的曲线积分计算对坐标曲线积分∫(6xy^2-y^3)dx+(6x^2y-3xy^2)dy为从点A(0,0)经曲线

答案:2.过程不详述了.这个积分是跟路径无关的,因为原函数是一个函数(3xxyy-xyyy)的全微分.在这种情况下,积分值等于原函数在起始点值的差.

(x-2xy)*(-xy+2y*y)-(3x*x-2xy)(x-9xy+6y*y)

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

全微分方程(3X^2+6xy^2)dx+(6x^2y+4y^2)dy=0的通解

(3X²+6xy²)dx+(6x²y+4y²)dy=03X²dx+4y²dy+(6xy²dx+6x²ydy)=0dx&s