Question

求下列微分方程的通解。2和4两个小题

Answer 1

解：（2）∵x(y^2-1)dx+y(x^2-1)dy=0
==>2xdx/(x^2-1)+2ydy/(y^2-1)=0
==>d(x^2-1)/(x^2-1)+d(y^2-1)/(y^2-1)=0
==>∫d(x^2-1)/(x^2-1)+∫d(y^2-1)/(y^2-1)=0
==>ln│x^2-1│+ln│y^2-1│=ln│C│ (C是常数)
==>(x^2-1)(y^2-1)=C
∴原方程的通解是(x^2-1)(y^2-1)=C；
（4）∵xdy+dx=e^ydx
==>xdy=(e^y-1)dx
==>dy/(e^y-1)=dx/x
==>e^(-y)dy/(1-e^(-y))=dx/x
==>d(1-e^(-y))/(1-e^(-y))=dx/x
==>∫d(1-e^(-y))/(1-e^(-y))=∫dx/x
==>ln│1-e^(-y)│=ln│x│+ln│C│ (C是常数)
==>1-e^(-y)=Cx
==>e^(-y)=1-Cx
∴原方程的通解是e^(-y)=1-Cx。