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Zhylliolom wrote:

/*But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise wink)

*/

I give up, would you give me a hint?

Zhylliolom wrote:

/*But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise wink)

*/

I give up, would you give me a hint?

**nefqu**- Replies: 0

Hi, this is my first post. I know that:

Sorry I don't know how to do derivatives in LaTeX.

Can someone help me prove the above statement?

These are two coupled equations, to solve it differentiate both sides, to get:

y''=x'=y

x''=y'=x

The general solution to this, where C and D are constants, is C sinh(t) and D cosh(t). Using your initial conditions:

x(0)=1

y(0)=0

x=cosh(t)

y=sinh(t).

Does this help?

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