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*...
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? 展开
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? 展开
2个回答
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