Need help with boundary conditions of a differential equation.

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