dy÷dx=x÷y的通解

设t=x/y则x=tydx=tdy+ydtdy/dx=y/(x+y^2)=>dx/dy=x/y+y把dx代入t+ydt/dy=t+yydt/dy=ydt/dy=1t=y+C(C是常数)x=y^2+Cy

分离变量法dy/y=(1+x)dx,两边积分,得ln|y|=x+x平方/2+C,整理得y=Ce的（x+x平方/2）方

dy/dx=(x+y)/(x-y)x+y=u,x-y=ty=(u-t)/2x=(u+t)/2dy/dx=(du+dt)/(du-dt)=u/tudu-udt=tdu+tdtudu-tdt=udt+td

特征方程r+1=0r=-1通解y=Ce^(-x)设特解y=axe^(-x)y'=ae^(-x)-axe^(-x)代入原方程得ae^(-x)-axe^(-x)+axe^(-x)=e^(-x)解得a=1因

令y/x=py=pxy'=p+p'x代入原方程得p+p'x=p+xp'x=xp'=1两边积分得p=x+Cy/x=x+C

dy/dx+y/x=cosx积分因子=e^∫1/xdx=e^ln|x|=x,乘以方程两边x·dy/dx+y=xcosxd(xy)/dx=xcosxxy=∫xcosxdxxy=∫xd(sinx)=xsi

dy/dx=-x/yydy=-xdx两边积分y²/2=-x²/2+Cy²=-2x²+Cy²+2x²=C再问：有错吧亲。y²/2=-

∫1/y*1/lnydy=∫1/sinxdxlnlny=∫1/2/[sin(x/2)*cos(x/2)]dxlnlny=ln(sin(x/2))-ln(cos(x/2))+clny=e^c*tan(x

dy/dx=1/(x+y)²令　　x+y=t原式变为　　d(t-x)/dx=1/t²即　　dt/dx=(1+t²)/t²变形得　　[t²/(1+t&#

∵dy/dx-2+y/x=0==>dy-2dx+ydx/x=0==>xdy-2xdx+ydx=0==>(xdy+ydx)-2xdx=0==>d(xy)-d(x^2)=0==>xy-x^2=C(C是常数

dy/dx-y/x=x运用公式y=e^(-∫-1/xdx)*[∫x*e^(∫-1/xdx)dx+C)]=x*(∫1dx+C)=x*(x+C)=x^2+Cx

答：dy/dx=y/(x+y)取倒数：dx/dy=(x+y)/y把x看做是y的函数,y是自变量x'(y)=(x+y)/yyx'=x+y(x/y)'=x'/y-x/y^2=(yx'-x)/y^2=y/y

没有检查,仅供参考.

dy/dx=-y/x分离变量1/xdx=-1/ydylnx=-lny整理得xy=c

设y=ux,dy/dx=u+xdu/dx原式化为u+xdu/dx=-1-udu/(1+2u)=-dx/x(1/2)ln|1+2u|=-ln|x|+lnC11+2y/x=C2/x^2x^2+2xy=C

两边同除以dx,整理后得到dy/dx=(x+y-1)/(x+y+1),然后转化一下,d(x+y)/dx=2(x+y)/(x+y+1).设u=x+y,得到du/dx=2u/(u+1).以下略.结果:x-

一阶线性常系数,可以有两种方法第一种,设函数u=u(x),与原式子相乘,使得等式左边=d(uy)/dxuy'+2uy=uxe^x由乘法法则可得du/dx=2udu/u=2dx∫du/u=∫2dxu=e

∵dy/dx=10^(x+y)==>10^(-y)*dy=10^x*dx==>∫10^(-y)*dy=∫10^x*dx==>-10^(-y)/ln10=10^x/ln10-C/ln10(C是积分常数)

dy÷dx=10^x乘10^ydy÷10^y=10^xdx-10^(-y)d(-y)=10^xdx两边积分-1/ln1010^(-y)=1/ln1010^x-10^(-y)=10^x10^x+10^(