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=02.利用函数的凹凸性,证明不等

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

d^2y/dx^2-dy/dx+ye^2x
=(d(dy/dx)/dt)(dt/dx)-(dy/dt)(dt/dx)+y(e^x)^2
=(d((dy/dt)(dt/dx))/dt)e^x-(dy/dt)e^x+yt^2
=(d((dy/dt)t)/dt)t-(dy/dt)t+yt^2
=(d^2y/dt^2)t+dy/dt-dy/dt+yt
=d^2y/dt^2+y=0

2楼正解－－！