how do you find the fundamental set of solutions for y"-4y=0? Any idea? The answer is in sines and cosines; how does that make sense?

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anonymous · Answer

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anonymous · Answer

The fundamental set of solutions for y"-4y=0 are not sines and cosines.
Sines and cosines are the fundamental set of solutions for y"+4y=0.
For the equation y''-4y=0 you can find the answer by thinking what functions f(x) have such second derivatives that they are almost equal to themselves
$$(f(x) \approx f''(x))$$
Exponential functions is what should come to mind (by knowing that (e^x)'=e^x). That is the only solution. Sines and cosines are just exponential functions taken with complex arguments (e^(ix)=cos(x)+i*sin(x)).
So if you figure that out try solving the equation with
$$y=Ae^{\lambda x}$$
Then find the constants A and lambda. (You will find that lambda can be 2 or -2 so the set of all solutions are functions e^2x and e^-2x)

anonymous · Answer

$$r^2-4=0$$
$$r=\pm2$$
you've got 2 real distinct roots, therefore:
$$y_c=c_1e^{2x}+c_2e^{-2x}$$