求方程(2x+y-4)dx+(x+y-1)dy=0的通解.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 01:33:54
![求方程(2x+y-4)dx+(x+y-1)dy=0的通解.](/uploads/image/z/5410238-14-8.jpg?t=%E6%B1%82%E6%96%B9%E7%A8%8B%282x%2By-4%29dx%2B%28x%2By-1%29dy%3D0%E7%9A%84%E9%80%9A%E8%A7%A3.)
x){ٴWtkUhWhThkX)glbb="}TِjՁePP-T%Dr*R*5qHѨ35NѨ37ҭ1+p뀨׆*֮u7RPp~6cݻ?ݱ
ص[Uj#|Աi|Hh46}0-.H̳- Y
求方程(2x+y-4)dx+(x+y-1)dy=0的通解.
求方程(2x+y-4)dx+(x+y-1)dy=0的通解.
求方程(2x+y-4)dx+(x+y-1)dy=0的通解.
∵(2x+y-4)dx+(x+y-1)dy=0
==>(2x-4)dx+(y-1)dy+(ydx+xdy)=0
==>d(x^2-4x)+d(y^2/2-y)+d(xy)=0
==>x^2-4x+y^2/2-y+xy=C/2 (C是任意常数)
==>2x^2-8x+y^2-2y+2xy=C
∴原方程的通解是2x^2-8x+y^2-2y+2xy=C.