求微分方程cosx(dy/dx)=ysinx+cos2x满足初始条件y∣x=π=1的特解．
微分方程的求解．dy/dx-(sinx/cosx)y=cox,y=e∫(sinx/cosx)y[C+∫cose-∫(sinx/cosx)dx]=(1/cosx){C+∫[(1+cos2x)/2]/dx}=(1/cos)(C+x/2+sin2x/4)由y∣x=π=1,得1=-C-π/2,即C=-1-π/2，故所求特解为y=(1/cosx)(x/2+sin2x/4-1-π/2)．
求微分方程cosx(dy/dx)=ysinx+cos2x满足初始条件y∣x=π=1的特解．
微分方程的求解．dy/dx-(sinx/cosx)y=cox,y=e∫(sinx/cosx)y[C+∫cose-∫(sinx/cosx)dx]=(1/cosx){C+∫[(1+cos2x)/2]/dx}=(1/cos)(C+x/2+sin2x/4)由y∣x=π=1,得1=-C-π/2,即C=-1-π/2，故所求特解为y=(1/cosx)(x/2+sin2x/4-1-π/2)．
