Trajectory of submarine chasing ship 
here's my try 
First I notice that if I can define L for every point then I define the curve and from the figure 
$$L^2 = x^2 + (y_1-y_2)^2$$
where $y_1$ is the corresponding y coordinate of $S_1$ and $Y_2$ is the y coordinate of the curve 
So using the fact that $ y'(x)=\tan\theta$ -where $\theta$ is the angel made with the positive direction of the X axis- is the slope of the tangent line to the curve at any point so an equation for L is $y=y'(x)X$ 
Now my whole idea is to find two different representations for L and equate them to each other to get an equation I can solve and I want here to use the arc length formula..so differentiating the equation of L we get
 $$y'= y''(x)X-y'(x)$$ 
now using the arc length formula 
$$\int\sqrt {1+\left(y''(x)X-y'(x)\right)^2}dx$$..I differentiate that with respect to x to get $\sqrt{1+\left(y''(x)X-y'(x)\right)^2}$ 
Now I  find another formula for the length of L which I get by using the fact that $(y_1-y_2) =-x\tan\theta = -y'(x)X$ and from the pythagorean theorem 
$$L =\sqrt {x^2 + (y'(x))^2x^2} = x\sqrt {1 + (y'(x))^2}$$ then differentiating this we get $$\sqrt {1 + (y'(x))^2} + \frac{xy''(x)}{\sqrt {1 + (y'(x))^2}}$$
then equating these two expressions for the derivative of the length of L we get
$$\sqrt{1+\left(y''(x)X-y'(x)\right)^2} = \sqrt {1 + (y'(x))^2} + \frac{xy''(x)}{\sqrt {1 + (y'(x))^2}}$$ which I can't solve any way I know even after a lot of simplification 
I know I didn't use the stated fact that they're moving at constant speeds which may simplify it a little bit but I can't seem to relate the two I also can't see how using time as the independent variable here helps 
one last thing..I want some problem solving advice on how to deal with situations like these not related to this specific problem..situations where the complexity of the model makes it impractical or unsolvable
Sorry I know this is quite long..Thanks in advance
 A: Hint.
Given $S_1(t) = (x_1(t),y_1(t))$ and $S_2(t) = (x_2(t),y_2(t))$
the pursuit law dictates that
$$
S_1(t) - S_2(t) = \lambda \dot S_2(t),\ \ \, \lambda > 0
$$
or normalizing
$$
\left< \frac{S_1(t)-S_2(t)}{|S_1(t)-S_2(t)|},\frac{\dot S_2(t)}{|S_2(t)|}\right > = 1
$$
where $\left< \cdot,\cdot \right>$ represents the scalar product of two vectors
now considering $S_1(t) = (0, v_1 t)$ and that $|\dot S_2| = v_2$
or
$$
v_2\sqrt{(y_2-t v_1)^2+x_2^2}=(t v_1-y_2) \dot y_2-x_2\dot x_2\\
\dot x_2^2+\dot y_2^2=v_2^2
$$
or
$$
x_2^2(\dot x_2^2-v_2^2)+2\dot x_2\dot y_2 x_2(y_2-v_1 t)+(v_1t-y_2)^2(\dot y_2^2-v_2^2) = 0\\
\dot x_2^2+\dot y_2^2=v_2^2
$$
and after substitution
$$
x_2^2\dot y_2^2-2x_2(v_1 t-y_2)\dot x_2\dot y_2 + (v_1 t-y_2)^2\dot x_2^2 = 0
$$
or
$$
(y_2 \dot x_2+x_2\dot y_2-v_1 t \dot x_2)^2=0\Rightarrow x_2 y_2'+y_2-v_1 t = 0
$$
here $y_2' = \frac{dy_2}{dx_2}$
we have also
$$
\dot x_2\sqrt{1+\dot y_2^2}=v_2 \Rightarrow dx_2\sqrt{1+\dot y_2^2}=v_2 dt\Rightarrow t = \frac{1}{v_2}\int \sqrt{1+y_2'^2}dx_2
$$
resulting in
$$
x_2 y_2'+y_2-\frac{v_1}{v_2}\int \sqrt{1+ y_2'^2}dx_2 = 0
$$
and after deriving
$$
x_2 y''_2 -\frac{v_1}{v_2}\sqrt{1+ y_2'^2} = 0
$$
now making $z = y'$ we have the ODE
$$
x_2 z' = \frac{v_1}{v_2}\sqrt{1+z^2}
$$
which is separable
$$
\frac{dz}{\frac{v_1}{v_2}\sqrt{1+z^2}} = \frac{d x_2}{x_2}
$$
etc.
A: Hint:
The position of $S_1$ is given by $$x_1=0,y_1=v_1t.$$
You know the speed
$$\frac{\sqrt{dx_2^2+dy_2^2}}{dt}=v_2$$
and direction of $S_2$,
$$\frac{dy_2}{dx_2}=\frac{y_2-v_1t}{x_2}.$$
