# Real Analysis Differentiation [closed]

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.

• Differentiate $g$ with respect to $x$ and factorize. – projectilemotion Jan 18 at 9:26
• Please don't link pictures, but write it directily here using MathJax. – mag Jan 18 at 9:31
• Please share your thoughts about the problem and let us know where you are stuck. Posting just problem statements is discouraged here. – Paramanand Singh Jan 18 at 11:00
• 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 – Anon Anon Jan 18 at 11:11
• sorry, my mistake @DavidC.Ullrich – Anon Anon Jan 18 at 11:52

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 $$g(x)=f(x)-(a\cos(x)+b\sin(x))$$.

• thank you very much! now I realized I miss interpreted the question – Anon Anon Jan 18 at 13:38
• @AnonAnon It's traditional for you to "accept" the answer by clicking on that arrow... – David C. Ullrich Jan 18 at 15:44

Solution of ode $$f+f''=0$$ is $$f=c_1\cos x+c_2\sin x$$. Then $$f^2+f'^2=c_1^2+c_2^2$$