Goodwill Problem Of Calculus

$\begin{aligned} f(x) & = 2018\left(1+x^2\right) \left(1+\int_0^x \frac {f(t)}{1+t^2} dt\right) \\ \frac {f(x)}{1+x^2} & = 2018 \left(1+\int_0^x \frac {f(t)}{1+t^2} dt\right) & \small \color{#3D99F6} \text{Let }g(x) = \frac {f(x)}{1+x^2} \\ g(x) & = 2018 \left(1+\int_0^x g(t)\ dt\right) & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ g'(x) & = 2018 g(x) \\ \implies g(x) & = {\color{#3D99F6}k} e^{2018x} & \small \color{#3D99F6} \text{where }k \text{ is a constant.} \end{aligned}$

Now we have:

$\begin{aligned} Q & = \frac 1{e^{2018}} \cdot \frac {f(2018)}{f(2017)} & \small \color{#3D99F6} \text{Note that }f(x) = (1+x^2)g(x) \\ & = \frac 1{e^{2018}} \cdot \frac {(1+2018^2)g(2018)}{(1+2017^2)g(2017)} \\ & = \frac {(1+2018^2)ke^{2018^2}}{(1+2017^2)ke^{2018\cdot 2017+2018}} \\ & = \frac {1+2018^2}{1+2017^2} \\ & \approx \boxed{1.000992} \end{aligned}$

Let $n = 2018$ , $g(x) = \frac{f(x)}{1 + x^2}$ , $f(x) = g(x) (1 + x^2)$ . Then we have:

$g(x) (1 + x^2) = n (1 + x^2) + n (1 + x^2) \int_0^x g(t) dt$ $g(x) = n + n \int_0^x g(t) dt$ Take the Laplace transform of both sides: $G(s) = \frac{n}{s} + n \frac{G(s)}{s}$ $G(s) = \frac{n}{s-n}$ And the inverse transform: $g(x) = n e^{n x}$

Therefore $f(x) = n e^{n x} (x^2+1)$ . Substituting values, the final answer is $814465/813658$ .

Rewrite the equation as $\int_0^x \frac{f(t)}{1 + t^2} dt = \frac{f(x)}{2018(1+x^2)} - 1$ Let's now define $g(x) = \int_0^x \frac{f(t)}{1 + t^2} dt$ . Because $f$ is a continuous function, $g(x)$ is a differentiable function and by the fundamental theorem of calculus, the previous equation is actually a differential equation on $g$ . Explicitly, $g = \frac{1}{2018}g' - 1$ . This is a first order linear equation which as we know has a general solution. We also have the initial condition $g(0) = \int_0^0 \frac{f(t)}{1 + t^2} dt = 0$ so this is enough to uniquely determine a solution. Skipping the known procedure of multiplying by an integrating factor and then integrating, the solution is $g(x) = e^{2018x} - 1$ .

But notice that $f(x) = 2018(1 + x^2)(g(x) + 1)$ so we now have found that $f(x) = 2018(1 + x^2)e^{2018x}$ .