Confusion on 1990 B1 Putnam question

Question

Find all real-valued continuously differentiable function $f$ on the real line such that for all $x$ $$(f(x))^2=\int_0^x(f(t))^2+(f’(t))^2\,\mathrm{d}t+1990$$

So the solution makes sense up to a certain point, we take the derivative of this on both sides and obtain $$2f(x)f’(x)=(f(x))^2+(f’(x))^2$$ Upon rearranging we find $$(f(x)-f’(x))^2=0$$ So $f(x)=Ce^x$. Since now we can solve for $c$ by seeing that $f(0)=\pm\sqrt{1990}$. My problem is that if we go back to the original with $f(x)=Ce^x$ then we get \begin{align*} (Ce^x)^2&=\int_0^x2(Ce^t)^2\,\mathrm{d}t+1990\\ &=C^2e^{2x}+1990 \end{align*} So it seems like $f(x)$ will never satisfy this because $0\neq1990$. Am I missing something here? Any value of $C$ wouldn’t do this as far as I can tell. Every solution says the same thing that $f(x)=\pm\sqrt{1990}e^x$ so what am I missing?

Answer 1

We have

\begin{align} \int_0^x((Ce^t)^2+(Ce^t)^2)dt+1990&=\int_0^x2C^2e^{2t}dt+1990\\ &=C^2e^{2x}-C^2+1990\\ &=C^2e^{2x} \end{align} while on the left $$(Ce^x)^2=C^2e^{2x}.$$

Answer 2

In evaluating the definite integral, you forgot to subtract the antiderivative evaluated at the lower bound of the integral.