求微分方程(x+sinx+siny)dx+cosydy=0通解如题,需要具体步骤

Question

求微分方程(x+sinx+siny)dx+cosydy=0通解
如题,需要具体步骤

Answer 1

∵(x+sinx+siny)dx+cosydy=0 ==>xe^xdx+e^xsinxdx+(e^xsinydx+e^xcosydy)=0
==>d(e^x(sinx-cosx))/2+d(e^x(x-1))+d(e^xsiny)=0
==>e^x(sinx-cosx)/2+e^x(x-1)+e^xsiny=C/2 (C是任意常数)
==>e^x(sinx-cosx)+2e^x(x-1)+2e^xsiny=C
==>e^x(sinx-cosx+2siny+2x-2)=C
∴原方程的通解是e^x(sinx-cosx+2siny+2x-2)=C.