作业帮求(y^2-3x^2)dy-2xydx=0的通解
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 09:59:47
sin(x^2+y^2)=x两边同时求导,得(x^2+y^2)'cos(x^2+y^2)=dx(2xdx+2ydy)cos(x^2+y^2)=dx2xdx+2ydy=dx/cos(x^2+y^2)2y
dy=d(tan(x^3)-2^-x)=sec²(x³)d(x³)+2^(-x)ln2dx=[3x²sec²(x³)+2^(-x)ln2]d
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
y'=2xsin4x-x²cos4x·4所以dy=(2xsin4x-4x²cos4x)dxy=ln√4＋t²=1/2ln(4+t²)y'=1/2·1/(4+t&
dy=d2x²++dxdy=2xdx+dx所以dy/dx=2x+1
y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy两边对x求导：dy/dx=f'[(x-1)/(x+1)]*2/（x+1)^2=arctan[(x-1)/(x+1)]
lny=xln(2+x)dlny=dxln(2+x)dy/y=ln(2-x)dx+x*1/(2+x)dxdy/(2+x)^x=[ln(2-x)+x/(2+x)]dxdy=(2+x)^x[ln(2-x)
y'=1/(x+x^2)*(2x+1)=(2x+1)/(x+x^2)dy=(2x+1)/(x+x^2)dx
y=x^2+3x+1dy=2x+3再问：��˼·再答：������
dy=2[e^x+e^(-x)]*[e^x-e^(-x)]dx再问：��������ϸ����再答：��������ϸ��������Dz��谡̫��û�취再问：������y���
x=sin(y/x)+e^2求dy/dxd(x)=d(sin(y/x)+e^2)dx=dsin(y/x)+de^2dx=cos(y/x)d(y/x)dx=cos(y/x)(xdy-ydx)/x^2x^
dy=2sin[x(x+1)]cos[x(x+1)](2x+1)
dy/dx=e^x/x^2-2e^x/x^3+2cos2x
求dy/dx=(x/y)+cos²(x/y)通解令x/y=u,则y=x/u,dy/dx=[u-x(du/dx)]/u²,代入原式得：[u-x(du/dx)]/u²=u+c
结果应该是C1x+C2+1/2y*log(x^2+y^2)+x*atan(y/x)希望采纳
dy/dx=(x-y+5)/(x+y-2)=[(x+3/2)-(y-7/2)]/[(x+3/2)+(y-7/2)]令v=y-7/2,u=x+3/2,原方程化为dv/du=(u-v)/(u+v)变为齐次
y=x^(2x)lny=2xlnx(1/y)dy=(2+2lnx)dxdy=x^(2x).(2+2lnx)dx
dy/d(x^3)=(dy/dx)/(d(x^3)/dx)=cosx/3(x^2)