Question

dy/dx=x/(y+x^2) 求解微分方程

Answer 1

x(dx/dy)-y-√(x^2+y^2)=0，除以y:(x/y)(dx/dy)-1-√((x/y)^2+1)=0令x/y=u，代入：u(u+yu')=√(u^2+1)+1yu'=(√(u^2+1)+1)/u-u=(√(u^2+1)+1-u^2)/uudu/(√(u^2+1)+1-u^2)=dy/ydu^2/(√(u^2+1)+1-u^2)=2dy/y积分∫dt/(√(t+1)+1-t)可令√(t+1)=z化成有理分式函数求解得：∫dt/(√(t+1)+1-t)=(-2/3)ln(√(t+1)+1)+(-4/3)ln(√(t+1)-2),代入得通解：(-2/3)ln(√(u^2+1)+1)+(-4/3)ln(√(u^2+1)-2)=2lny+(-2/3)lnC或通解：(√(u^2+1)(√(u^2+1)-2)^2=C/y^3其中：u=x/y