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Real analysis question about differentiation

Assume that $f$ and $f'$ are differentiable on $\mathbf R$ and that for every x in $\mathbf R$, $f(x) + f''(x) = 0 $. Show that $g (x)= f^2(x) + (f'(x))^2$ is a constant.

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  • $\begingroup$ Differentiate $g$ with respect to $x$ and factorize. $\endgroup$ – projectilemotion Jan 18 at 9:26
  • $\begingroup$ Please don't link pictures, but write it directily here using MathJax. $\endgroup$ – mag Jan 18 at 9:31
  • $\begingroup$ Please share your thoughts about the problem and let us know where you are stuck. Posting just problem statements is discouraged here. $\endgroup$ – Paramanand Singh Jan 18 at 11:00
  • $\begingroup$ I really don't understand which to prove first, in order to prove g(x) is constant, g'(x) must be equals to zero for all x in the intervals, and I've tried to prove the f(x)+f''(x)=0 using the Rolle's theorem and now I'm confused @ParamanandSingh $\endgroup$ – Anon Anon Jan 18 at 11:11
  • $\begingroup$ sorry, my mistake @DavidC.Ullrich $\endgroup$ – Anon Anon Jan 18 at 11:52
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Suppose that $f:\Bbb R\to\Bbb R$ and $f+f''=0$.

Then $g=f^2+(f')^2$ is constant.

Proof: $g'=2ff'+2f''f'=2f'(f+f'')=0$.

It's amusing to note that this proves uniqueness for the solution:

Cor. If $f(0)=f'(0)=0$ then $f=0$.

And hence

Cor. If $f''+f=0$, $f(0)=a$ and $f'(0)=b$ then $f(x)=a\cos(x)+b\sin(x)$.

Proof: The previous corollary shows that $g=0$, if $g(x)=f(x)-(a\cos(x)+b\sin(x))$.

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  • $\begingroup$ thank you very much! now I realized I miss interpreted the question $\endgroup$ – Anon Anon Jan 18 at 13:38
  • $\begingroup$ @AnonAnon It's traditional for you to "accept" the answer by clicking on that arrow... $\endgroup$ – David C. Ullrich Jan 18 at 15:44
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Solution of ode $f+f''=0$ is $f=c_1\cos x+c_2\sin x$. Then $$f^2+f'^2=c_1^2+c_2^2$$

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