∵齊次方程y'+ysinx=0 ==>y'=-ysinx
==>dy/y=-sinxdx
==>ln│y│=cosx+ln│C│ (C是積分常數(shù))
==>y=Ce^cosx
∴齊次方程是y=Ce^cosx (C是積分常數(shù))
于是,設(shè)原微分方程的通解是y=C(x)e^cosx (C(x)表示關(guān)于x的函數(shù))
∵y'=C'(x)e^cosx-C(x)sinxe^cosx
代入原方程,得C'(x)e^cosx=e^cosx
==>C'(x)=1
==>C(x)=x+C (C是積分常數(shù))
∴y=C(x)e^cosx=(x+C)e^cosx
故原微分方程的通解是y=(x+C)e^cosx (C是積分常數(shù))