2道高数的题
2道高数的题
1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=0
2.利用函数的凹凸性,证明不等式：sin(x/2)>x/π（0
dy/dx=(dy/dt)/(dt/dx)=t(dy/dt)
d^2y/dx^2-dy/dx+ye^2x=0
d(dy/dx)/dx-dy/dx+ye^2x=0
d(t(dy/dt))/dx-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+t(dy/dt)-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+yt^2=0
(d^2y/dt^2)+y=0
2,设f(x)=sin(x/2)-x/π,（0
设x=λ*0+（1-λ)*π,(0那么根据上凸函数的性质,有
f(x)=f(λ*0+（1-λ)*π)>λf(0)+(1-λ)f(π)=0
因此sin(x/2)-x/π>0,（0
1,X=lnt,那么dx=(1/t)dt,dt/dx=t
dy/dx=(dy/dt)/(dt/dx)=t(dy/dt)
d^2y/dx^2-dy/dx+ye^2x=0
d(dy/dx)/dx-dy/dx+ye^2x=0
d(t(dy/dt))/dx-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+t(dy/dt)-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+yt^2=0
(d^2y/dt^2)+y=0
2,设f(x)=sin(x/2)-x/π,（0
设x=λ*0+（1-λ)*π,(0那么根据上凸函数的性质,有
f(x)=f(λ*0+（1-λ)*π)>λf(0)+(1-λ)f(π)=0
因此sin(x/2)-x/π>0,（0