1.∵y=C1sin(x-C2)
∴y'=C1cos(x-C2)
∵ x=π時,y'=0.即C1cos(π-C2)=0
∴cos(π-C2)=0 （∵x=π時,y=1.∴C1≠0）
==>π-C2=kπ+π/2(k=0,±1,±2,±3,.)
==>C2=π/2-kπ
∵x=π時,y=1.即C1sin(π-C2)=1
∴C1sin(kπ+π/2)=1 ==>C1cos(kπ)=1
==>C1*(-1)^k=1
==>C1=1,k=2n (n=0,±1,±2,±3,.)
∴所求特解是y=1*sin(x-π/2+2nπ)
=sin(x-π/2)
=-sin(π/2-x)
=-cosx.
2.∵ln|y|=y/x-C1 ==>y=e^(y/x-C1)
==>y=e^(y/x)*e^(-C1)
令C=e^(-C1)
∴通解是y=Ce^(y/x) (C是積分常數(shù))