Question

求解微分方程x''-1/t*x'+(x')^2=0

大神求教啊

Answer 1

解：令t=e^z，则tx'=dx/dz，t²x''=d²x/dz²-dx/dz
代入原方程，化简得
d²x/dz²-2dx/dz+(dx/dz)²=0..........(1)
令dx/dz=p，则d²x/dz²=dp/dz
代入方程(1)，化简得[1/p+1/(2-p)]dp=2dz
==>ln│p│-ln│2-p│=2z+ln│C1│ (C1是积分常数)
==>p/(2-p)=C1e^(2z)
==>p=2[C1e^(2z)-1]/[C1e^(2z)+1]
==>dx=2[C1e^(2z)-1]dz/[C1e^(2z)+1]
==>x=ln│[C1e^(2z)+1][e^(-2z)+C1]│+ln│C2│ (C2是积分常数)
==>e^x=C2(C1t²+1)(1/t²+C1)
==>e^x=C2(C1t+1/t)²
故 原方程的通解是e^x=C2(C1t+1/t)²。