由方程组x=t^2+2t和t^2-y+asiny=1(0＜a＜1) 确定y为x的函数,求d^2y/dx^2
问题描述：
由方程组x=t^2+2t和t^2-y+asiny=1(0＜a＜1) 确定y为x的函数,求d^2y/dx^2
答
由方程组x=t²+2t和t²-y+asiny=1(0＜a＜1) 确定y为x的函数,求d²y/dx²
dy/dx=(dy/dt)/(dx/dt)
设F(t,y)=t²-y+asiny-1≡0；则dy/dt=-(∂F/∂t)/(∂F/∂y)=(2t)/(1-acosy)；dx/dt=2t+2；
∴dy/dx=y′=[(2t)/(1-acosy)]/(2t+2)=t / [(t+1)(1-acosy)]
故d²y/dx²=(dy′/dt)/(dx/dt)
其中dy′/dt={[(t+1)(1-acosy)]-t[(1-acosy)+(t+1)(asiny)(dy/dt)]}/[(t+1)(1-acosy)]²
={1-acosy-at(t+1)siny[(2t)/(1-acosy)]}/[(t+1)(1-acosy)]²
=[(1-2cosy)²-2at²(t+1)siny]/[(t+1)²(1-acosy)³]
∴d²y/dx²=[(1-2cosy)²-2at²(t+1)siny]/[2(t+1)³(1-acosy)³]