微分方程dx+xydy＝y^2dx

(x^2+y^2)dx-xydy=0dy/dx=(x²+y²)/(xy)dy/dx=((x/y)²+1)/(x/y)令u=y/x则dy=du*x+dx*udy/dx=(d

少了括号两边积分∫y/(y^2-1)dy＝∫1/(x-1)dx得1/2×ln(y^2-1)＝ln(x-1)＋1/2lnC等式的前两部分的对数都没有加绝对值,所以常数项用lnC,一是为了容易消去对数运算

若题目为xydy/dx=y^2+x^2,则：dy/dx=x/y+y/x令y/x=u,则：y=xu,dy/dx=u+xdu/dx=x/y+y/x=u+1/uu+xdu/dx=u+1/uxdu/dx=1/

这是一阶齐次微分方程(x^2+y^2)dx-xydy=0dy/dx=(x²+y²)/(xy)dy/dx=((x/y)²+1)/(x/y)令u=y/x则dy=du*x+dx

令：u=y/x则：y=xudy/dx=u+xdu/dx由：(x^2+y^2)dx=xydydy/dx=(x^2+y^2)/xy=x/y+1/[x/y]dy/dx=u+xdu/dx=u+1/uxdu/d

(x-2y^2)dx+2xydy=0M=x-2y^2N=2xyM_y=-4yN_x=2y(M_y-N_x)/N=-6y/2xy=-3/x假设有关于x的积分因子u(Mu)_y=(Nu)_xM_y*u=N

方程是可分离变量的ydy/(y²+1)=-dx/x两边积分得：∫y/(y²+1)dy=-∫1/xdx得：(1/2)∫1/(y²+1)d(y²)=-ln|x|ln

设u=y/xdy/dx=(x+y²)/xyu+xdu/dx=1/xu+uudu=1/x²dx1/2u²=-1/x+Cy²=2Cx²-2x希望能帮道楼主

（x+2y）dx+ydy＝0,设y=tx,则dy=xdt+tdx,化为dx/x=-tdt/(t+1)^2=[-1/(t+1)+1/(t+1)^2]dt,lnx+c'=-ln(t+1)-1/(t+1),

dy/dx=-(x^2+2xy)/(xy)=-(x+2y)/y=-x/y-2令u=y/x,则y=xu,y'=u+xu'代入原方程：u+xu'=-1/u-2xu'=-1/u-2-u=-(u+1)^2/u

应该是可分离变量的吧两端同除以x^2得（1+(y/x)^2)dx=2y/xdy令y/x=uy=uxy'=u'x+u上式变为(1+u^2)=2u(u'x+u)整　理1+u^2=2u^2+2uu'x1-u

方程变形为dy/dx=x/y+y/x.令u=y/x,则y=xu,dy/dx=u++x*du/dx,所以原方程化为u+x*du/dx=u+1/u.所以udu=dx/x.两边积分1/2*u^2=lnx+l

由(x^2+y^2)dx-2xydy=0得到dy/dx=(x^2+y^2)/2xy=0.5(x/y+y/x)设y/x=z,则y=zxdy/dx=xdz/dx+z=0.5(1/z+z)化为zdz/(1-

设P(x,y)＝x^2＋y^2,Q(x,y)＝2xy,则αP/αy＝αQ/αx,所以此微分方程是全微分方程（x^2+y^2)dx+2xydy=0x^2dx＋(y^2dx+2xydy)＝0d(x^3/3

都是非线性非齐次微分方程

原式两边乘以x^2得x^2e^xdx+3x^2y^2dx+2x^3ydy=0x^2e^xdx+dx^3y^2=0x^2e^xdx=-dx^3y^2两边积分得∫x^2e^xdx=-∫dx^3y^2x^2

令：y2=xu，则：2ydy=xdu+udx，则原微分方程可化为：（xu+x）dx-x（xdu+udx）=0，即：xdx-x2du=0，所以：dx-xdu=0即：du＝dxx，解得：u=ln|x|+c

(1)dx+xydy=y=y^2dx+ydy==>(xy-y)dy=(y^2-1)dx==>(x-1)ydy=(y^2-1)dx==>ydy/(y^2-1)=dx/(x-1)两边积分,得：ln(y^2

这是一阶齐次微分方程(x^2+y^2)dx-xydy=0dy/dx=(x²+y²)/(xy)dy/dx=((x/y)²+1)/(x/y)令u=y/x则dy=du*x+dx