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。