Question

cosxsinydx十sinxcosydy=0求解

Answer 1

1.解：∵cosxsinydx+sinxcosydy=0
==>d(sinxsiny)=0
==>sinxsiny=C (C是常数)
∴原方程的通解是sinxsiny=C；
2.解：∵xy'=y+√(x^2+y^2)
==>xdy-ydx=√(x^2+y^2)dx
==>(xdy-ydx)/x^2=√(1+(y/x)^2)dx/x (等式两端同除x^2)
==>d(y/x)=√(1+(y/x)^2)dx/x
==>d(y/x)/√(1+(y/x)^2)=dx/x
==>ln│y/x+√(1+(y/x)^2)│=ln│x│+ln│C│ (C是常数)
==>y/x+√(1+(y/x)^2)=Cx
==>y+√(x^2+y^2)=Cx^2
∴原方程的通解是y+√(x^2+y^2)=Cx^2；
3.解：∵dy/dx=3xy+xy^2
==>dy/dx=xy(y+3)
==>dy/(y(y+3))=xdx
==>[1/y-1/(y+3)]dy=3xdx
==>ln│y│-ln│y+3│=3x^2/2+ln│C│ (C是常数)
==>y/(y+3)=Ce^(3x^2/2)
==>y=C(y+3)e^(3x^2/2)
∴原方程的通解是y=C(y+3)e^(3x^2/2)。