Question

求(2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0的通解 (2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0

Answer 1

∵(2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0
==>(ydx+xdy)+(2xy^2dx+x+2x^2y)-x^4y^3dy=0
==>d(xy)+d((xy)^2)-x^4y^3dy=0
==>-d(xy)-d((xy)^2)+x^4y^3dy=0
==>-d(xy)/(xy)^4-d((xy)^2)/((xy)^2)^2+dy/y=0 (等式两端同除(xy)^4)
==>(1/3)/(xy)^3+1/(xy)^2+ln│y│=ln│C│ (等式两端积分，C是常数)
==>(xy+1/3)/(xy)^3+ln│y│=ln│C│
==>ye^((xy+1/3)/(xy)^3)=C
∴原方程的通解是ye^((xy+1/3)/(xy)^3)=C。