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1.E^x+yDy=Dx 2.Dy/Dx=y*x+y 3.y'+yCosx=E^%sinx,y...

1.e^(x+y)dy=dxe^ydy=e^(-x)dxe^y+C=-e^(-x)e^y+e^(-x)+C=02.dy/y=(x+1)dxln(y/C)=x+x^2/2y=Ce^(x+x^2/2)3.令y=C(x)e^(-sinx)代入得dC(x)/dx=1C(x)=x+cy(0)=3 C(0)=3 c=3y=(x+3)e^(-sinx)

1.e^xdx+ydy=dx (e^x-1)dx+ydy=0通解e^x-x+(1/2)y^2=c2.dy/dx=y*x+y dy/dx=y(x+1)dy/y=(x+1)dx通解lny+(1/2)(x+1)^2+c3.y'+ycosx=e^-sinx,y(0)=3dy+ycosxdx=e^(-sinx)dxdy+yd(sinx)=e^(-sinx)dxe^sinxdy+ye^(sinx)d(sinx)=dxe^sinxdy+yde^sinx=dxdye^(sinx)=dx通解ye^sinx=x+c

两边同时乘以e^sinx得到:e^sinxy'+e^sinxcosxy=1即:(e^sinxy)'=1∴e^sinxy=x+C∴通解为:y=(x+C)e^(-sinx)

解:∵e^(y^2+x)dx+ydy=0 ==>e^(y^2)*e^xdx=-ydy ==>-2ye^(-y^2)dy=2e^xdx ==>e^(-y^2)d(-y^2)=2e^xdx ==>e^(-y^2)=2e^x+C (C是常数) ∴原方程的通解是e^(-y^2)=2e^x+C.

解:∵(e^(x+y))dx+ydy=0 ==>e^xdx+ye^(-y)dy=0 ==>e^x+(1-y)e^(-y)=C (C是常数) ∴原方程的通解是e^x+(1-y)e^(-y)=C.

x'=dx/dy=xy+x^2y^3,同除以x^2得--x'/x^2+y/x+y^3=0,即 d(1/x)/dy+y(1/x)+y^3=0.令1/x=u 于是u'+yu+y^3=0,通解为 u=--2(y^2/2--1)+Ce^(--y^2/2).即1/x=Ce^(--y^2/2)+2--y^2.

由x^3*(dy/dx)=x^2*y-1/2*y^3可得:(dy/dx)=y/x-1/2*(y/x)^3 ……①设y/x=U(x),则y=u*x那么dy/dx=du/dx *x +u此时①式即:du/dx *x +u=u-(1/2)* u^3所以:x*du/dx=-(1/2)* u^3 当u≠0有dx/x= -2* du/{ u^3}的lnx +c=1/[u^2] =(x/y)^2带入y|(x=1)=1得c=1带入c整理一下答案就出来了.

1、ydy-e^(y^2+3x)dx=0ydy=e^(y^2+3x)dxydy/e^(y^2)=e^(3x)dx两边积分得1/2ln[e^(y^2)]=1/3e^(3x)+C 1/2y^2=1/3e^(3x)+C2、y(1+x^2)dy-x(1+y^2)dx=0ydy/(1+y^2)=xdx/(1+x^2)两边积分得1/2ln(1+y^2)=1/2ln(1+x^2)+C得

y^2dx+ydy=x^2ydy-dx (y^2-1)dx=y(x^2-1)dy dx/(x^2-1)=ydy/(y^2-1) 即0.5[(1/(x-1)-1/(x+1)]dx=0.5dy^2/(y^2-1) 两边积分得: ln|(x-1)/(x+1)|=ln|y^2-1| 即|y^2-1|=(x-1)/(x+1) 哪里不清欢迎追问,满意谢谢采纳!

原式=> ydy=(x^2+y^2-x)dx 令x^2+y^2=t>=0 则两边分别微分得:2xdx+2ydy=dt 故原式=> dt-2xdx=2(t-x)dx=> dt/2t=dx 所以 lnt*1/2=x+C 所以原方程解为 ln(x^2+y^2)=2x+C

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