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高等数学之微分方程求解
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后面的指数看不清:
设函数f(x,y)可微,∂f/∂x=-f(x,y),
f(0,π/2)=1,
且lim(h->0)[f(0,y+h)/f(0,y)]^(1/h)=e^cosy,求f(x,y)
是e^cosy吗?
f(0,y)是纯粹的y的函数,我们可以先求出这个函数。
h->0,1/h->∞,极限是e为底的指数函数,lim(h->0)[f(0,y+h)/f(0,y)=0必然成立,
设g(h)=[f(0,y+h)/f(0,y)]^(1/h)
取对数:
lng(h)=[lnf(0,y+h)-lnf(0,y]/h
0/0型,用洛必达法则,或根据导数定义:
lim(h-->0)lng(h)=lim(h-->0)[lnf(0,y+h)-lnf(0,y]/h=d(lnf(0,y))/dy=cosy
lnf(0,y)=siny+C,f(0,y)=e^[siny+C]=De^siny
f(0,π/2)=1代入:
f(0,y)=De^sin(π/2)=De=1,D=1/e
f(0,y)=e^(siny-1)
∂f/∂x=-f(x,y),把f(x,y)仅仅看成x的函数:
∂f/f=-1dx
积分:lnf=-x+C,
f=De^(-x),变系数法。
D看成(x,y)的函数,D应该仅仅是y的函数,
∂f/∂x=∂D/∂x.e^(-x)-De^(-x),∂D/∂x=0,
设f(x,y)=D(y)e^(-x)
∂f/∂y=e^(-x)dD(y)/dy
∂f/∂y|(0,π/2)=f'y(0,y)=cosye^(siny-1)=dD(y)/dy
D(y)=e^(siny-1)
∴f(x,y)=e^(siny-1)e^(-x)
验证:
∂f/∂x=-e^(siny-1)e^(-x)=-f(x,y),正确:
f(0,π/2)=e^(sin(π/2)-1).e^(-0)=1,正确;
lim(h-->0)[f(0,y+h)/f(0,y)]^(1/h)
=lim(h-->0)[e^(sin(y+h)-1)/e^(siny-1)]^(1/h)
=lim(h-->0)[e^(sin(y+h)--siny)]^(1/h)
=lim(h-->0)e^{[sin(y+h)--siny]/h}
=e^sin'y
=e^cosy
正确:
设函数f(x,y)可微,∂f/∂x=-f(x,y),
f(0,π/2)=1,
且lim(h->0)[f(0,y+h)/f(0,y)]^(1/h)=e^cosy,求f(x,y)
是e^cosy吗?
f(0,y)是纯粹的y的函数,我们可以先求出这个函数。
h->0,1/h->∞,极限是e为底的指数函数,lim(h->0)[f(0,y+h)/f(0,y)=0必然成立,
设g(h)=[f(0,y+h)/f(0,y)]^(1/h)
取对数:
lng(h)=[lnf(0,y+h)-lnf(0,y]/h
0/0型,用洛必达法则,或根据导数定义:
lim(h-->0)lng(h)=lim(h-->0)[lnf(0,y+h)-lnf(0,y]/h=d(lnf(0,y))/dy=cosy
lnf(0,y)=siny+C,f(0,y)=e^[siny+C]=De^siny
f(0,π/2)=1代入:
f(0,y)=De^sin(π/2)=De=1,D=1/e
f(0,y)=e^(siny-1)
∂f/∂x=-f(x,y),把f(x,y)仅仅看成x的函数:
∂f/f=-1dx
积分:lnf=-x+C,
f=De^(-x),变系数法。
D看成(x,y)的函数,D应该仅仅是y的函数,
∂f/∂x=∂D/∂x.e^(-x)-De^(-x),∂D/∂x=0,
设f(x,y)=D(y)e^(-x)
∂f/∂y=e^(-x)dD(y)/dy
∂f/∂y|(0,π/2)=f'y(0,y)=cosye^(siny-1)=dD(y)/dy
D(y)=e^(siny-1)
∴f(x,y)=e^(siny-1)e^(-x)
验证:
∂f/∂x=-e^(siny-1)e^(-x)=-f(x,y),正确:
f(0,π/2)=e^(sin(π/2)-1).e^(-0)=1,正确;
lim(h-->0)[f(0,y+h)/f(0,y)]^(1/h)
=lim(h-->0)[e^(sin(y+h)-1)/e^(siny-1)]^(1/h)
=lim(h-->0)[e^(sin(y+h)--siny)]^(1/h)
=lim(h-->0)e^{[sin(y+h)--siny]/h}
=e^sin'y
=e^cosy
正确:
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