How to find a function that is defined by its integral and its derivative?

Question

I was trying to solve a physics problem. The question is: "How long will it take for a boat to cross a river if its velocity is always directed towards a fixed point on the opposite side of the river bank, assuming the velocity of the boat is greater than the water's velocity?". I came to the conclusion that it would be easier to describe the boat's movement if we took the point towards which the boat is directed as moving at a constant velocity equal to the current's velocity. Thinking about the problem this way, the function describing the boat's $y$ position would satisfy: $$f'\left(x\right)=\left(\frac{1}{d-x}\right)\ast\left(\frac{v_{river}}{v_{boat}}\ast\int_{0}^{x}{f\left(x\right)\ dx}-f(x)\right)$$ Where $d$ is the width of the river. We also know that $f(0) = 0$, $f'(0)=0$, and $F(0)=0$. Is there any way to deduce $f(x)$ knowing this information? Or is there a simpler way to think of the problem in the first place?

Answer 1

As I already mentioned within the comments your notation is a little bit confusing. By assuming the two occuring $x$ within the integral are not the same - at by setting $k=v_{river}/v_{boat}$ we will get

$$\begin{align} f'(x)&=\left(\frac{1}{d-x}\right)\left(k\int_{0}^{x}f(t) dt-f(x)\right)\\ f'(x)&=\left(\frac{1}{d-x}\right)\left(k[F(x)-F(0)]-f(x)\right)\\ f'(x)&=\left(\frac{1}{d-x}\right)(kF(x)-f(x))\\ (d-x)f'(x)&=kF(x)-f(x) \end{align}$$

Now differentiate the whole equation w.r.t. $x$ yields to

$$\begin{align} (d-x)f''(x)-f'(x)&=f(x)-f'(x)\\ f''(x)&=\frac k{d-x} f(x) \end{align}$$

The solution of this differential equation is - according to WolframAlpha - is given in terms of the modified Bessel Functions

$$f(x)=c_1\sqrt{k(d-x)}I_1\left(2\sqrt{k(d-x)}\right)+c_2\sqrt{k(d-x)}K_1\left(2\sqrt{k(d-x)}\right)$$

where $I_n(x)$ denotes the modified Bessel Function of the first kind and $K_n(x)$ the modified Bessel Function of the second kind respectively.