Need help with boundary conditions of a differential equation. QUESTION:

A particle $A$ is moving along the $X$ axis at a constant horizontal velocity $u\hat{i}$. Another particle $B$ is moving such that its velocity vector always points towards the particle $A$. $B$ moves with a constant speed $v$. At time $t = 0$, position of $A$ is $(0,0)$ and that of $B$ is $(0, L)$. What is the time when the particles collide?

What I did:

I set up the differential equations, however I am getting something stupid. This is what I did:

Let the position vector of $A$ be 
$$\vec{a}(t) = ut\hat{i}$$

and that of $B$ be $$\vec{b}(t) = x(t)\hat{i} + y(t)\hat{j} = x\hat{i} + y\hat{j}$$ (for notational convenience).

Now, the velocity vector of $B$ is 
$$\vec{v}(t) = \lambda(\vec{a}(t) - \vec{b}(t))$$

where we'll $\lambda$ is some constant that we'll figure out later. This equation is true as $\vec{v}$ is collinear with the vector $\vec{BA}$.

So, 
$$ \dfrac{d}{dt} \Big(\vec{b}(t)\Big) = \lambda(\vec{a}(t) - \vec{b}(t))$$
$$ \dfrac{d}{dt} \Big(x\hat{i} + y\hat{j}\Big) = \lambda((ut-x)\hat{i} - y\hat{j})$$

Separating the components (is this step wrong? I don't know vector calculus, but I believe this must be true) we have:

$$ \dfrac{d}{dt}(x) = \lambda\cdot(ut-x) $$
$$ \dfrac{d}{dt}(y) = \lambda\cdot(-y) $$

I could solve the first differential equation. Solving the second one:
$$ \dfrac{dy}{y} = - \lambda \cdot dt $$
That part was easy but the boundary conditions are the trouble makers:

$$ \text{Intitial: } y = L, t = 0 \\ \text{Final: } y = 0, t = T $$

So,
$$ \ln y\Bigg|_{L}^{0} = -\lambda \cdot t \Bigg|_{0}^T \,\,\,\, \text{Ln(0)?!?!?!} $$

What's wrong?

Or is there any other way to solve the problem? Then Hints are of course welcome :D
$$a(t) = ut i$$
let particle $B$ make an angle $\theta$ with y axis .then you will see 
$$B(t) =vtsin\theta i  + (L - tvcos\theta) j$$
hen they collide $a(T) = b(T)$
T is time of collision .we get $$sin\theta = \frac{u}{v}..(1)$$
and$$cos\theta = \frac{L}{vt}...(2)$$
square $(1)$ and $(2)$ and add , further solve for $T$.
we get,
$$T = \frac{L}{ \sqrt{v^2 - u^2} }$$
$$a(t) = ut i$$
let particle $B$ make an angle $\theta$ with y axis .then you will see 
$$B(t) =vtsin\theta i  + (L - tvcos\theta) j$$
hen they collide $a(T) = b(T)$
T is time of collision .we get $$sin\theta = \frac{u}{v}..(1)$$
and$$cos\theta = \frac{L}{vt}...(2)$$
square $(1)$ and $(2)$ and add , further solve for $T$.
we get,
$$T = \frac{L}{ \sqrt{v^2 - u^2} }$$