d+2y+dx+2如何计算

万烁步3065求方程y^2+x^3(dy/dx)=xy(dy/dx)的通解
包宁勇18413326768 ______ y^2+x^3(dy/dx)=xy(dy/dx)即(y^2)dx+(x^3)dy=(xy)dy 即2(y^2)dx+2(x^3)dy=(2xy)dy=xd(y^2)即2(y^2)dx+2(x^3)dy=xd(y^2) 即2x(y^2)dx+2(x^4)dy=(x^2)d(y^2)即(y^2)d(x^2)+2(x^4)dy=(x^2)d(y^2) 即2(x^4)dy=(x^2)d(y^2)-(y^2)d(x^2)即2dy=[(x^2...

万烁步3065dy/dx=2x^3+3xy^2+x/3x^2y+2y^3 - y的通解? -
包宁勇18413326768 ______ 提供一下思路,具体结果自己算算.dy/dx=(2x^3+3xy^2+x)/(3x^2y+2y^3-y),dy/dx=x(2x^2+3y^2+1)/y(3x^2+2y^2-1),ydy/xdx=(2x^2+3y^2+1)/(3x^2+2y^2-1),dy^2/dx^2=(2x^2+3y^2+1)/(3x^2+2y^2-1).dy^2/dx^2=[2(x^2-1)+3(y^2+1)]/[3(x^2-1)+2(y^2+1)].设:u=(...

万烁步3065∫(x^(1/2))/((x^(1/2) - 1)dx
包宁勇18413326768 ______ 设√x=y,x=y².∫[√x/(√x-1)]dx=∫[y/(y-1)]dy²=2∫y²/(y-1)dy=∫{2y+2)dy+[2/(y-1)]d(y-1)} =y²+2y+2ln(y-1)=x+2√x+2ln(√x-1)

万烁步3065求微分方程(xy^2 - x)dx+(x^2y+y)dy=0的通解 -
包宁勇18413326768 ______ (xy^2-x)dx+(x^2y+y)dy=0 xy^2dx-xdx+x^2ydy+ydy=0 xy^2dx+x^2ydy-xdx+ydy=02xy^2dx+2x^2ydy-2xdx+2ydy=0 注意:d(x^2y^2)=2xy^2dx+2x^2ydy 所以:d(x^2y^2)-2xdx+2ydy=0 通解为:x^2y^2-x^2+y^2=C 也可以写成(x^2+1)(y^2-1)=C

万烁步3065常微分方程dy/dx=x+y/x - y -
包宁勇18413326768 ______ dy/dx=(x+y)/(x-y) x+y=u,x-y=t y=(u-t)/2 x=(u+t)/2 dy/dx=(du+dt)/(du-dt)=u/t udu-udt=tdu+tdt udu-tdt=udt+tdu d(u^2-t^2)=2dut u^2-t^2=2ut+C(x+y)^2-(x-y)^2=2(x+y)(x-y)+C2x*2y=2(x^2-y^2)+C2xy=(x^2-y^2)+C'

万烁步3065求文档: 求全微分(x^2+2xy)dx+xydy=0的通解 -
包宁勇18413326768 ______ (x^2+2xy)dx+xydy=0 x^2dx+2xydx+xydy=0(x/y)dx+2dx+dy=0 x/y=u dx=ydu+udy uydu+u^2dy+2ydu+2udy+dy=0(uy+2y)du+(u^2+2u+1)dy=0-(u+2)du/(u+1)^2=dy/y-du/(u+1)-du/(u+1)^2=dlny d-ln(u+1)+1/(u+1)=dlny dlny+ln(u+1)-1/(u+1)=0 lny+ln(u+1)-1/(u+1)=C lny(u+1)-1/(u+1)=C ln(x+y)-y/(x+y)=C

万烁步3065作变量代换x=lnt简化方程d^2y/dx^2 - dy/dx+e^2x*y=0 -
包宁勇18413326768 ______ 解:x=lnt dx/dt=1/t dy/dx=(dy/dt)/(dx/dt)=t dy/dt d²y/dx²=[d/dt(dy/dx)]/(dx/dt)=t² d²y/dt²+t dy/dt 代入d^2y/dx^2-dy/dx+e^2x*y=0 t² d²y/dt²+t dy/dt-t dy/dt+e^(2lnt)y=0 t² d²y/dt²+t²y=0 d²y/dt²+y=0 希望对你有帮助,望采纳,谢谢~

万烁步3065常微分方程 解方程(dy/dx)3 - 3y2dy/dx=0 -
包宁勇18413326768 ______ 令dy/dx=p,则有p³-3py²=p(p²-3y²)=0 故得p=dy/dx=0,y₁=C₁为一个解;由p²-3y²=0得 p²=3y²;故p=±(√3)y; 分离变量得:dy/y=±(√3)dx 积分之得:lny=±(√3)x+lnC ∴y₂=Ce^(±x√3)为另一个解.

万烁步3065y＂=(1+y'^2)y' -
包宁勇18413326768 ______ 求微分方程 y''=(1+y'²)y' 的通解 解:令y'=dy/dx=p,则y''=dy'/dx=dp/dx=(dp/dy)(dy/dx)=pdp/dy;代入原式得:pdp/dy=(1+p²)p;由此可知:p=y'=0,即y=C是方程的一个解;消去p得:dp/dy=1+p²; 分离变量得:dp/(1+p²)=dy;积分之得:arctanp=y+c₁;即p=dy/dx=tan(y+c₁);故 x=∫cot(y+c₁)dy=∫cot(y+c₁)d(y+c₁)=ln∣sin(y+c₁)∣+ln(1/c₂)=ln(∣sin(y+c₁)∣/c₂); 即 ∣sin(y+c₁)∣=c₂e^x是原方程的通解.

万烁步3065求解微分方程:(x^2+y^2+2x)dx+2ydy=0 麻烦给出过程,x+ln(x^2+y^2)=C -
包宁勇18413326768 ______[答案] (x^2+y^2+2x)dx+2ydy=0 (x^2+y^2)dx+d(x^2)+d(y^2)=0 (x^2+y^2)dx+d(x^2+y^2)=0