求微积分齐次方程通解求齐次方程(1+2e^(x/y))dx+2(1-x/y)e^(x/y)dy=0的通解

Question

求微积分齐次方程通解
求齐次方程(1+2e^(x/y))dx+2(1-x/y)e^(x/y)dy=0的通解

Answer 1

∵令x=yu，则dx=ydu+udy
代入原方程，化简得 y(1+2e^u)du+(u+2e^u)dy=0
==>yd(u+2e^u)+(u+2e^u)dy=0
==>d(y(u+2e^u))=0
==>∫d(y(u+2e^u))=0
==>y(u+2e^u)=C (C是积分常数)
==>y(x/y+2e^(x/y))=C
==>x+2ye^(x/y)=C
∴原方程的通解是x+2ye^(x/y)=C。