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求微分方程cosx(dy/dx)=ysinx+cos²x满足初始条件y∣_x=π=1的特解．

高老师2年前 (2024-03-25)高等数学(一)(00020)5

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)．

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求微分方程cosx(dy/dx)=ysinx+cos2x满足初始条件y∣x=π=1的特解．

求微分方程cosx(dy/dx)=ysinx+cos²x满足初始条件y∣_x=π=1的特解．