y³dx (2xy²-1)dy=0

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 02:31:20
y³dx (2xy²-1)dy=0
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

求解微分方程 x^2*dy/dx=xy-y^2

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^(

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

xy+e的平方+y=2 ,求dy/dx

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

微分方程(1+y^2)dx+(xy-genhao1+y^2 cosy)dy=0

∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0(等式两端同除√(1+y

∫(0→1)dy∫(0→y)根号下(y^2-xy)dx=

这是我的解答,希望对你有帮助,有疑问请追问,若满意还望采纳,祝生活愉快!

dy/dx=1+x+y^2+xy^2

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

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

解微分方程y^2+(x^2)(dy/dx)=xy(dy/dx)

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^

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

求解微分方程(1-2xy)dy/dx=y(y-1),

方程有关于y的积分因子再问:能写下过程吗?万分感谢再答:

微分方程 xy-1/x^2y dx - 1/xy^2 dy =0

是xy-[1/(x^2y)]dx-[1/(xy^2)]dy=0还是[(xy-1)/(x^2y)]dx-[1/(xy^2)]dy=0请表达清楚,无歧义!再问:[(xy-1)/(x^2y)]dx-[1/(

xy-sin(πy^2)=0 求dy/dx

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

(1+y^2)dx+(xy-根号下(1+y^2 ) cosy)dy=0

∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0(等式两端同除√(1+y

求齐次微分方程dy/dx=y^2/xy-x^2

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

dy/dx=xy/x^2-y^2

你要求什么?