求(2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0的通解

ssdpowerw2022-10-04 11:39:541条回答

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

已提交,审核后显示!提交回复

共1条回复
划过天际的弧 共回答了24个问题 | 采纳率95.8%
∵(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。
1年前

相关推荐