$$\large \text{Problem}$$
I'll give the question first and then the motivation. Is there a closed form solution to the differential equation
$$\sqrt{k^2-{f'(t)}^2}f(t)-\sqrt{t^2-{f(t)}^2}f'(t)=t\sqrt{k^2-1}$$ with initial value $$f{\left(\frac{1}{\sqrt{k^2-1}}\right)}=\frac{1}{\sqrt{k^2-1}}?$$
What about $$\sqrt{k^2-{g'(t)}^2}(1-g(t))+\sqrt{t^2-{(1-g(t))}^2}g'(t)=t\sqrt{k^2-1}$$ with initial value $$g{\left(\frac{1}{\sqrt{k^2-1}}\right)}=1?$$
Alternatively, can we give the solution to one as a function of the solution to the other? These are increasingly tricky approaches to solving the problem given in the motivation.
$$\large \text{Motivation}$$ I have been playing in my spare time with what happens when we have two "blobs" in $\mathbb{R}^2$ that radially expand from every point in the blob at a constant speed, starting as two points, but stop expanding anywhere they come into contact. More formally, we have sets $A_t$ and $B_t$ indexed by time $t\geq0$ such that $A_0=(0,0)$ and $B_0(1,0)$. For all $x\in\mathbb{R}^2$ and times $s>t$, if there exists a path from $A_t$ to $x$ of length $k(s-t)$ that does not intersect $B_r$ for any $r<s$, the $x\in A_s$. For all $x\in\mathbb{R}^2$ and times $s>t$, if there exists a path from $A_t$ to $x$ of length $s-t$ that does not intersect $B_r$ for any $r<s$, the $x\in A_s$. The difference being the appearance of $k$, the speed of $A_t$'s expansion relative to that of $B_t$. We suppose that $k>1$, without loss of generality. For all $x\in A_s$, there must be a path $\gamma(t)$ to $x$ such that $\gamma(t)$'s length changes at a rate of at most $k$, and for all $t<s$, $\gamma(t)\in A_t\setminus B_t$. Similarly for $B_s$, but we change the rate of growth from $k$ to $1$. For a point $x\in\mathbb{R}^2$, if there exists $t$ such that $x\in A_t\cap B_t$, then for all times $s$, if $x\in A_s$, then $x\in B_s$. These conditions uniquely define $A_t$ and $B_t$. I am interested in the final shapes of $A=\bigcup_{t\geq0}A_t$, and $B=\bigcup_{t\geq0}B_t$. We can show that $B$ is a closed set, and $A$ is the closure of its complement, so we can obviously just focus on $B$.
I know that $B$ is convex and bounded, and its boundary can be split into three curves. The first is easy. We consider the $x$ such that the distance from $x$ to $(0,0)$ is $k$ times the distance from $x$ to $(1,0)$. If there is no issue with the straight line path between $x$ and $(0,0)$ intersecting with $B_t$, or the straight line path between $x$ and $(1,0)$ intersecting with $A_t$, then the paths clearly belong to $A$ and $B$ respectively, and they meet at these boundary points. The set of $x$ satisfying this distance equation is a certain circle, with $(1,0)$ on the inside and $(0,0)$ on the outside. Things are okay until the path from $(0,0)$ becomes tangent to the circle, at which point it's no longer obvious that it avoids $B_t$, and indeed, it does not. There are two points on the circle which meet these tangent paths. These are the beginnings of the other two curves, which by symmetry, will obviously be mirror images in the line $y=0$. So we might as well focus on the top.
We can show that what needs to happen is that as we trace along the curve at speed $k$, we just meet up with the expansion of $A_t$. Now, we can show without much difficulty that $B_t$ is only ever going to expand via straight lines from $(1,0)$. These lines are going to be growing radially from the points on the boundary of a circle of radius $t$ centred at $(1,0)$ which have not already been obstructed by the growth of $A_t$. To get a little heuristic, locally, at a point $c(t)$ on the curve, we're going to be meeting an expanding wall of growth at unit speed, going up in parallel lines. What kind of motion is going to beat the wall? Well, we'll be moving at speed $k$, so we need to pick an angle such that the movement in the direction of the wall is at unit speed. This is actually already satisfied by the derivative of the first curve at the starting point of $c$. The curves will stop when they run into each other somewhere along $y=0$.
$$\text{To summarise}$$ the magnitude of $c$'s derivative is $k$. $c$ is always on a point of the circle of radius $t$ centred at $(1,0)$. $c$'s derivative forms a constant angle with the tangent line to this circle, which we can calculate from the initial value. We can also calculate the initial value of $t$, where $A_t$ and $B_t$ no longer grow in lines colliding with each other, and the curve starts. Writing out the coordinates of $c$ and its derivative, and putting all this into the language of inner products, we find that we can solve for the $x$-coordinates from the $y$-coordinates and $t$ and vice versa. We then put the equations together to get the $y$ coordinate given by the function $f(t)$ above. $g(t)$ gives the $x$ coordinate.
$$\text{EDIT}$$
It's been pointed out to me that the properties defining my differential equation are satisfied by logarithmic spirals (by the differential equation, this means that logarithmic spirals with given initial conditions uniquely satisfy them). Logarithmic spirals are however given in polar coordinates, so I suppose my original question becomes one of expressing a part of a logarithmic spiral as a closed-form Cartesian function. But I can't find this, so I suspect that there is none. Anyway, as far as characterising the set goes I'm satisfied. It's more natural to shift everything to the left by $1$. Then the boundary is given by these curves:
- The part of circle with centre $\left(\frac{1}{k^2-1},0\right)$ and radius $\frac{k}{k^2-1}$ to the left of the $y$-axis,
- The polar curve $r=\frac{1}{\sqrt{k^2-1}}e^\frac{2\theta-3\pi}{2\sqrt{k^2-1}}$ from $\theta=\frac{3\pi}{2}$ to $2\pi$, and
- The reflection of that curve in the $x$-axis.
Here is an obligatory picture of boundaries for a family of $B$'s for different $k$'s, which is certainly what I wanted out of this.
Edit: answer is invalid after OP changed the problem.
I am sorry if I understand your question wrong. To me it seems there is no solution to the ODE you gave. If we manipulate the ODE like this: $$ \sqrt{k^2-{f'(t)}^2}f(t)-\sqrt{t^2-{f(t)}^2}f'(t)=\frac{k^2t}{\sqrt{k^2-1}}\\ \Rightarrow\sqrt{k^2-{f'(t)}^2}f(t)=\sqrt{t^2-{f(t)}^2}f'(t)+\frac{k^2t}{\sqrt{k^2-1}}\\ \Rightarrow(k^2-{f'(t)}^2)f(t)^2=\left(\sqrt{t^2-{f(t)}^2}f'(t)+\frac{k^2t}{\sqrt{k^2-1}}\right)^2\\ \Rightarrow f(t)^2k^2-\frac{k^4t^2}{k^2-1}-f'(t)^2t^2 = 2k^2tf'(t)\frac{\sqrt{t^2-f(t)^2}}{\sqrt{k^2-1}}\\ \Rightarrow \left(f(t)^2k^2-\frac{k^4t^2}{k^2-1}-f'(t)^2t^2 \right)^2= \left(2k^2tf'(t)\frac{\sqrt{t^2-f(t)^2}}{\sqrt{k^2-1}}\right)^2\\ \Rightarrow f'(t)^2=-\frac{k^2}{(k^2-1)t^2}\left((f(t)^2-t^2)k^2+f(t)^2\pm 2\sqrt{f(t)^2k^2(f(t)^2-t^2)}\right) $$ However from we the original equation we need $t^2\geq f(t)^2$ so there exists no solution unless $f(t)^2=t^2$, which does not solve the ODE for all $k$.