做边量替换,u=y/x,即y=ux
y’=u+xu'
原方程左右同除x^2y
变为(1-u+u^2)+(1/u+1+u)(u+xu')=0
积分再换回变量就是答案了
不知道你会不会积分,还是写下过程吧,没算出来积分化简可得(1+u+u^2)/[u(1+u^2)](du)=-2(dx)/x[1/u+1/(1+u^2)](du)=-2(dx)/xln|u|+arctanu=-2lnx+c所以ln(|u|*x^2)=c-arctanu(y/x)*x^2=e^[c-arctan(y/x)]xy=ce^[-arctan(y/x)](这里c是上式e^c)
解微分方程y(x^2-xy+y^2)+x(x^2+xy+y^2)dy/dx=0
解微分方程y(x^2-xy+y^2)+x(x^2+xy+y^2)dy/dx=0
答案yx=ce^[-arctan(y/x)]
答案yx=ce^[-arctan(y/x)]
数学人气:619 ℃时间:2019-09-22 07:47:28
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