1.∵齊次方程y'+y/x=0 ==>dy/y=-dx/x
==>ln│y│=ln│C│-ln│x│ (C是積分常數(shù))
==>y=C/x
∴設(shè)原方程的通解是y=C(x)/x (C(x)是關(guān)于x的函數(shù))
∵y'=[C'(x)x-C(x)]/x²
代入原方程,得[C'(x)x-C(x)]/x²+C(x)/x²=sinx/x
==>C'(x)=sinx
==>C(x)=-cosx+C (C是積分常數(shù))
∴原方程的通解是y=(C-cosx)/x
∵當x=π時y=1 ==>(C+1)/π=1 ==>C=π-1
∴滿足條件的解是y=(π-1-cosx)/x
2.∵齊次方程y'+ycotx=0 ==>dy/y=-cosxdx/sinx
==>dy/y=-d(sinx)/sinx
==>ln│y│=ln│C│-ln│sinx│ (C是積分常數(shù))
==>y=C/sinx
∴設(shè)原方程的通解是y=C(x)/sinx (C(x)是關(guān)于x的函數(shù))
∵y'=[C'(x)sinx-C(x)cosx]/sin²x
代入原方程,得[C'(x)sinx-C(x)cosx]/sin²x+[C(x)/sinx]cotx=5e^cosx
==>C'(x)=5sinx*e^cosx
==>C(x)=-5∫e^cosxd(cosx)
==>C(x)=-5e^cosx+C (C是積分常數(shù))
∴原方程的通解是y=(C-5e^cosx)/sinx
∵當x=π/2時,y=-4 ==>C-5=-4 ==>C=1
∴滿足條件的解是y=(1-5e^cosx)/sinx
==>ysinx+5e^cosx=1