(3y-7x+3)dx+(7y-3x+3)dy=0求通解

Question

(3y-7x+3)dx+(7y-3x+3)dy=0求通解

Answer 1

令x=m+3/10,y=n-3/10,则dx=dm,dy=dn
代入原方程,化简得 (3n-7m)dm+(7n-3m)dn=0.(1)
令n=mt,则dn=tdm+mdt
代入方程(1),化简得7dm/m=(3-7t)dt/(t^2-1)
==>7dm/m=[-2/(t-1)-5/(t+1)]dt
==>7ln│m│=-2ln│t-1│-5ln│t+1│+ln│C│ (C是非零常数)
==>m^7=C/[(t-1)^2*(t+1)^5]
==>m^7=C/[(n/m-1)^2*(n/m+1)^5]
==>m^7*(n/m-1)^2*(n/m+1)^5=C
==>(n-m)^2*(n+m)^5=C
==>(y-x+3/5)^2*(y+x)^5=C
故原方程的通解是(y-x+3/5)^2*(y+x)^5=C.