Possible Error in the 1990 B1 Putnam Answer Key

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.

• Good questions (+1) Mar 8, 2021 at 4:15
• 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. Mar 8, 2021 at 4:18
• Something is wrong here. Left side is a function of $x$, whilst the right side is a function of $a$. Mar 8, 2021 at 4:20
• yes, @user58697, in the link the left side is $f(a)^2$ Mar 8, 2021 at 4:22
• Thank you, I'll correct my typo. Guess I'm a bit of a hypocrite, heh. Mar 8, 2021 at 4:31

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