matlab求解微分方程的通解问题解某微分方程：x^2*D2y+x*Dy+(x^2-1/2)*y=0,初值：y(pi/2)=2,Dy(pi/2)=-2/pi,书本标准答案为：ans=2^(1/2)*pi*^(1/2)/x^(1/2)*sin(x)；但我运行：dsolve('x^2*D2y+x*Dy+(x^2-1/2)*y=0','y(pi/2)=2,Dy(p

　　matlab求解微分方程的通解问题

　　解某微分方程：x^2*D2y+x*Dy+(x^2-1/2)*y=0,初值：y(pi/2)=2,Dy(pi/2)=-2/pi,书本标准答案为：

　　ans=2^(1/2)*pi*^(1/2)/x^(1/2)*sin(x)；

　　但我运行：dsolve('x^2*D2y+x*Dy+(x^2-1/2)*y=0','y(pi/2)=2,Dy(pi/2)=-2/pi','x')

　　ans=

　　(2*bessely(-2^(1/2)/2,x)*(2^(1/2)*besselj(-2^(1/2)/2,pi/2)-besselj(-2^(1/2)/2,pi/2)+pi*besselj(1-2^(1/2)/2,pi/2)))/(pi*(besselj(1-2^(1/2)/2,pi/2)*bessely(-2^(1/2)/2,pi/2)-bessely(1-2^(1/2)/2,pi/2)*besselj(-2^(1/2)/2,pi/2)))-(2*besselj(-2^(1/2)/2,x)*(2^(1/2)*bessely(-2^(1/2)/2,pi/2)-bessely(-2^(1/2)/2,pi/2)+pi*bessely(1-2^(1/2)/2,pi/2)))/(pi*(besselj(1-2^(1/2)/2,pi/2)*bessely(-2^(1/2)/2,pi/2)-bessely(1-2^(1/2)/2,pi/2)*besselj(-2^(1/2)/2,pi/2)))

　　即使我simpify(ans),j结果还是一样,怎么回事啊?