Question

求微分方程满足初始条件的特解：y''=e^2y,y(0)=y'(0)=0

Answer 1

令p=y'=dy/dt,那么有：
y''=dp/dt=(dp/dy)(dy/dt)=pdp/dy
将上述结果代入原方程得到：
p(dp/dy)=exp(2y)
分离变量得到：
pdp=exp(2y)dy
等式两侧取不定积分得到：
p²/2=[exp(2y)]/2+M···········································M为任意常数
整理得到：y'=dy/dt=p=√[exp(2y)+N]·····································N为任意常数,N=2M
代入初始条件解得：N=-1
再次分离变量得到：
dy/√[exp(2y)-1]=dt····················································※·
令y=lnθ,即θ=exp(y),等式两边取微分得到：dy=dθ/θ
将以上结果代入方程※,得到：
dθ/[θ√(θ²-1)]=dt······················································
利用换元积分法,令θ=secδ,同理等式两边取微分得到：dθ=secδtanδdδ,代入方程,得到：
dδ=dt
两边积分得到：
δ=t+L···········································L为任意常数
所以有：
exp(y)=θ=secδ=sec(t+L)
代入初始条件得到：L=0
所以有题目方程的特exp(y)=sect 或者y=ln|sect|