Question

做到这不知道怎么做了，帮忙解一下这个微分方程嘛，谢谢

Answer 1

dt/dx = -cost
dt/cost = -dx
sect dt = - dx
∫sectdt
=∫dt/cost
=∫costdt/cos²t
=∫dsint/cos²t
=∫dsint/(1-sin²t)
=(1/2)[∫dsint/(sint+1)-∫dsint/(sint-1)]
=(1/2)(ln|sint+1|-ln|sint-1|)+C
=(1/2)ln|(sint+1)/(sint-1)|+C (对数里分子分母都乘以sint+1)
=(1/2)ln|(sint+1)²/cos²t|+C
=ln|(sint+1)/cost|+C
=ln|tant+sect|+C
所以
ln|tant+sect|=-x+C1
tant+sect=Ce^(-x)
所以tan(y-x)+sec(y-x)=Ce^(-x)

Answer 2

t'=dt/dx
移项把t放左边，x放右边就积出来了