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x=sin(t), dx/dt=cos(t); y=cos(t), dy/dt=-sin(t).

It follows that x²+y²=1 by the definition of trig functions.




Let u=dy/dx=-tan(t), then d²y/dx²=du/dx.



d²y/dx²=du/dx=-sec³(t), d²y/dx²+sec³(t)=0.

But sec³(t)≠4, because t is a variable parameter, not a constant.

Now let’s work in the reverse direction: 

d²y/dx²=du/dx=-4, u=-4x+a, where a is a constant.

u=dy/dx=-4x+a, y=-2x²+ax+b, where b is another constant.

Substitute for x and y:

cos(t)=-2sin²(t)+asin(t)+b, which is not generally true, because a and b are arbitrary constants.

If we apply the special case a=b=0, y=-2x²=-2sin²(t)≠cos(t)=y (given) in general. y≠y is clearly false.

Or, 1-2sin²(t)≡cos(2t)≠1+cos(t) except in certain cases.

Whichever way we look at this, x=sin(t), y=cos(t) does not imply that d²y/dx²+4=0. So the question is invalid.

by Top Rated User (796k points)

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