求二阶微分方程的通解y''*e^y'=1

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求二阶微分方程的通解
y''*e^y'=1

令y'(x)=t(x) y''(x)=t'(x)
原式=t'(x) * e^t=1
t'(x)=e^-t
t(x)=-e^-t + C1
t+e^-t=C1
y'+e^-y'=c1
y+e^-y=c1x+c2

∵y''*e^y'=1 ==>e^y'd(y')=dx
==>e^y'=x+C1 (C1是积分常数)
==>y'=ln│x+C1│
==>y=∫ln│x+C1│dx
==>y=xln│x+C1│-∫[x/(x+C1)]dx

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∵y''*e^y'=1 ==>e^y'd(y')=dx
==>e^y'=x+C1 (C1是积分常数)
==>y'=ln│x+C1│
==>y=∫ln│x+C1│dx
==>y=xln│x+C1│-∫[x/(x+C1)]dx
==>y=xln│x+C1│-∫[1-C1/(x+C1)]dx
==>y=xln│x+C1│-x+C1ln│x+C1│+C2 (C2是积分常数)
==>y=(x+C1)ln│x+C1│-x+C2
∴原方程的通解是y=(x+C1)ln│x+C1│-x+C2 (C1,C2是积分常数)

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