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Find all real differentiable functions $f$ that satisfy the following.

$$f(a)^2 = \int_0^a (f(x)^2 + f'(x)^2)dx + 1990 \forall a \in \mathbb{R}$$

The solution manual is found here.

Putting $y = f(x)$ and differentiating the relationship gives $2y y' = y^2 + y'^2$ or $(y - y')^2 = 0$. So $y = y'$ or $y = -y'$. Integrating, $y = Ae^x$ or $y = - Ae^x$. But $y(0) = \pm \sqrt{1990}$, so $f(x) = \pm \sqrt{1990}e^x$.

Wouldn't the differential linear differential equation $y'=-y$ yield a solution of $y=Ae^{-x}$ as opposed to $y=-Ae^x$? Additionally, I am unsure why the solution $y=-y'$ is even necessary.

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  • $\begingroup$ Good questions (+1) $\endgroup$
    – J. W. Tanner
    Mar 8, 2021 at 4:15
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    $\begingroup$ I don't think that solutions page is from a publication of the MAA. It appears to be the work of one author, so some mistakes could be expected. $\endgroup$
    – aschepler
    Mar 8, 2021 at 4:18
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    $\begingroup$ Something is wrong here. Left side is a function of $x$, whilst the right side is a function of $a$. $\endgroup$
    – user58697
    Mar 8, 2021 at 4:20
  • $\begingroup$ yes, @user58697, in the link the left side is $f(a)^2$ $\endgroup$
    – J. W. Tanner
    Mar 8, 2021 at 4:22
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    $\begingroup$ Thank you, I'll correct my typo. Guess I'm a bit of a hypocrite, heh. $\endgroup$
    – Talmsmen
    Mar 8, 2021 at 4:31

1 Answer 1

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The procedure is correct, but you only get $y'=y$ (which follows from $(y'-y)^2=0$), hence $y=Ce^x$ and, setting $a=0$, $y=\pm\sqrt{1990}e^x$

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