You have some wrong signs in the last two equations.
Based on the equation of the derivative of the pursuit curve $y=f(x)$ described by the chaser object that you indicate
$$
\frac{dy}{dx}=\frac{ut-y}{p-x}\tag{1}
$$
I assume that the chased object moves along the straight line $x=p$, as I commented above. Assume further that it moves upwards. (See remark 2). Then
$$
t=\frac{y}{u}+\frac{p-x}{u}\frac{dy}{dx}.\tag{2}
$$
and from
$$
s=\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =vt\tag{3}
$$
we conclude that
$$
t=\frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi .\tag{4}
$$
Equating the two equations for $t$ $(2)$ and $(4)$ we get
$$
\frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =\frac{y}{u}+
\frac{p-x}{u}f(x).
$$
Differentiating both sides we obtain the equation (note that the LHS is
positive)
$$
\frac{1}{v}\sqrt{1+(f^{\prime }(x))^{2}}=\frac{p-x}{u}f^{\prime \prime }(x).\tag{5}
$$
If we let $w=\frac{dw}{dx}=f^{\prime }(x)$ this equation corresponds to the following one in $w$ and $w^{\prime }=\frac{dw}{dx}$
$$
\sqrt{1+w^{2}}=k(p-x)\frac{dw}{dx}\qquad w=f^{\prime }(x),k=\frac{v}{u},\tag{6}
$$
which can be rewritten as
$$
\frac{dw}{\sqrt{1+w^{2}}}=\frac{1}{k}\frac{dx}{p-x}\tag{7}
$$
by applying the method of separation of variables to $x$ and $w$. The integration is easy
$$
\begin{eqnarray*}
\int \frac{dw}{\sqrt{1+w^{2}}} &=&\int \frac{1}{k}\frac{dx}{p-x}+\log C \\
\text{arcsinh }w &=&-\frac{1}{k}\log \left( p-x\right) +\log C.
\end{eqnarray*}
$$
The initial condition $x=0,w=f^{\prime }(0)=0$ yields
$$
\begin{eqnarray*}
-\frac{1}{k}\log p+\log C &=&\text{arcsinh }0=0 \\
&\Rightarrow &\log C =\frac{1}{k}\log p ,
\end{eqnarray*}
$$
which means that
$$
\text{arcsinh }w=-\frac{1}{k}\log \left( p-x\right) +\frac{1}{k}\log p=-\frac{1}{k}\log \frac{p-x}{p}.\tag{8}
$$
Solving for $w$ we get
$$
\begin{eqnarray*}
\frac{dy}{dx} &=&w=\sinh \left( -\frac{1}{k}\log \frac{p-x}{p}\right) =\frac{1}{2}\left( e^{-\frac{1}{k}\log \frac{p-x}{p}}-e^{\frac{1}{k}\log \frac{p-x}{p}}\right) \\
&=&\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1/k}-\frac{1}{2}\left( \frac{p-x}{p}\right) ^{1/k}\tag{9}
\end{eqnarray*}
$$
To integrate this equation consider two cases.
- (a) $k=\frac{v}{u}>1$
$$\begin{eqnarray*}
y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1/k}-\left( \frac{p-x}{p}\right) ^{1/k} dx \\
&=&-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+C.
\end{eqnarray*}$$
The constant of integration $C$ is defined by the initial condition $x=0,y=0$
$$\begin{eqnarray*}
0 &=&-\frac{1}{2}\frac{pk}{k-1}+\frac{1}{2}\frac{pk}{k+1}+C \\
&\Rightarrow &C=\frac{pk}{k^{2}-1}.
\end{eqnarray*}$$
Hence
$$y=-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+\frac{pk}{k^{2}-1}$$
$$\tag{10}$$
The chaser overtakes the chased object at the point $(p,f(p))$, with $f(p)=
\frac{pk}{k^{2}-1}$.
- (b) $k=\frac{v}{u}=1$. We have
$$\frac{dy}{dx}=\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1}-\frac{1}{2}\left(
\frac{p-x}{p}\right)
$$
and
$$\begin{eqnarray*}
y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1}-\left( \frac{
p-x}{p}\right) dx \\
&=&-\frac{1}{2}p\ln \left( p-x\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}+C.
\end{eqnarray*}$$
The same initial condition $x=0,y=0$ yields now
$$\begin{eqnarray*}
C &=&\frac{1}{2}p\ln \left( p\right) \\
&& \\
y &=&-\frac{1}{2}p\ln \left( \frac{p-x}{p}\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}.\tag{11}
\end{eqnarray*}$$
The chaser never overtakes the chased object.
Example for (a): graph of $y=f(x)$ for $k=2,p=50$
Example for (b): graph of $y=f(x)$ for $k=1,p=50$
Remarks:
This answer is similar to the answer of mine to the question Cat Dog problem using integration.
It was inspired by Helmut Knaust's The Curve of Pursuit.
