求微分方程的通解 y''=e^y
5个回答
令P=y',则
y''=dP/dx=(dP/dy)·(dy/dx)= P dP/dy
则
P dP/dy =e^y
→P dP =e^y·dy
两边积分得
(1/2)P² =e^y +(1/2)C1²;
P² =2e^y +C1²
P=√(2e^y +C1²)
则
dy/dx=√(2e^y +C1²)
→dx=(1/√(2e^y +C1²) ) dy
积分得
x=∫1/√(2e^y +C1²) dy
=∫e^y/[e^y·√(2e^y +C1²)] dy
=∫1/[e^y·√(2e^y +C1²)] d e^y
令e^y=t,则
x=∫1/[t·√(2t +C1²)] d t
再令u=√(2t +C1²),则t=(u²-C1²)/2
dt=udu
则x=∫u / [u(u²-C1²)/2] d u
=2∫1 / (u²-C1²) d u
=(1/C1)·∫[1/(u-C1) - 1/(u+C1)] d u
=(1/C1)· ln|(u-C1)/(u+C1)| +C2
=(1/C1)· ln|(√(2t +C1²)-C1) / (√(2t +C1²)+C1)| +C2
=(1/C1)· ln|(√(2e^y +C1²) - C1) / (√(2e^y +C1²) + C1)| + C2
原微分方程的解是
x=(1/C1)· ln|(√(2e^y +C1²) - C1) / (√(2e^y +C1²) + C1)| + C2
