博文
改进的生存方程求解方法
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非常感谢陈小斌博主提示，这里改进一下对方程（3）求解。
[begin{array}{l}
frac{1}{B}frac{{dB}}{{dt}} = a - {b_1}{A_1} - {b_2}{A_2} \
d(frac{{dB}}{{Bdt}})/dt = - {b_1}frac{{d{A_1}}}{{dt}} - {b_2}frac{{d{A_2}}}{{dt}} \
- {B^{ - 2}}{(frac{{dB}}{{dt}})^2} + {B^{ - 1}}frac{{{d^2}B}}{{d{t^2}}} = - {b_1}({c_1}B - {e_1}t) - {b_2}({c_2}B - {e_2}t) \
{B^{ - 1}}frac{{{d^2}B}}{{d{t^2}}} - {B^{ - 2}}{(frac{{dB}}{{dt}})^2} + ({b_1}{c_1} + {b_2}{c_2})B - ({b_1}{e_1} + {b_2}{e_2})t = 0 \
Bfrac{{{d^2}B}}{{d{t^2}}} - {(frac{{dB}}{{dt}})^2} + ({b_1}{c_1} + {b_2}{c_2}){B^3} - ({b_1}{e_1} + {b_2}{e_2}){B^2}t = 0\
end{array}]
[X = frac{{dB}}{{dt}}]
则：
方程（3）变为：
[begin{array}{l}
BXfrac{{dX}}{{dB}} - {X^2} + ({b_1}{c_1} + {b_2}{c_2}){B^3} - ({b_1}{e_1} + {b_2}{e_2}){B^2}t = 0 \
X = BY \
frac{{d{Y^2}}}{{dB}} + 2({b_1}{c_1} + {b_2}{c_2}) - 2({b_1}{e_1} + {b_2}{e_2}){B^{ - 1}}t = 0 \
{Y^2} = Z - 2({b_1}{c_1} + {b_2}{c_2})B \
frac{{dZ}}{{dB}} - 2({b_1}{e_1} + {b_2}{e_2}){B^{ - 1}}t = 0 \
end{array}]
方程得到简化，可以得到更准确的尝试解。
[begin{array}{l}
BXfrac{{dX}}{{dB}} - {X^2} + ({b_1}{c_1} + {b_2}{c_2}){B^3} - ({b_1}{e_1} + {b_2}{e_2}){B^2}t = 0 \
X = BY \
frac{{d{Y^2}}}{{dB}} + 2({b_1}{c_1} + {b_2}{c_2}) - 2({b_1}{e_1} + {b_2}{e_2}){B^{ - 1}}t = 0 \
{Y^2} = Z - 2({b_1}{c_1} + {b_2}{c_2})B \
frac{{dZ}}{{dB}} - 2({b_1}{e_1} + {b_2}{e_2}){B^{ - 1}}t = 0 \
end{array}]
方程得到简化，可以得到更准确的尝试解。
https://blog.sciencenet.cn/blog-361477-570636.html
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