Partial Differentials I've just started learning about PDE's and am struggling with this question.

Consider the equation
  $$\cos^2(x)  \frac{\partial{u}}{\partial x} + y \frac{\partial{u}}{\partial y} =0$$
  Find the characteristic curves in the explicit form. Hence obtain the general solution of the PDE. For each of the following boundary conditions, find the particular solution or explain why it does not exist:
  
  
*
  
*(a) $u(0, y) = y^2$.
  
*(b) $u(x, 0) = x^2$.
  

So far I have done this but am unsure of how to complete the question:
$\dfrac{dy}{dx} = \dfrac{y}{\cos^2(x)}$
$\int_y\dfrac{dy}{y} =  \int_ x\dfrac{dx}{\cos^2(x)}$
$\ln(y)= \tan(x) +C$
$y= C e^{\tan(x)} $
$C=y e^{-\tan(x)} $
$u=f(e^{-\tan(x)}) $
 A: I claim that (a) is solvable and (b) is a characteristic problem. Okay, we have the PDE
$$\cos^2(x)  \frac{\partial{u}}{\partial x} + y \frac{\partial{u}}{\partial y} =0$$
with initial conditions
$$\begin{cases} \text{(a) } ~ u(0, y) = y^2 \\ \text{(b) } ~
u(x, 0) = x^2 \end{cases}$$
We do this with the method of characteristics. There is a nice determinant criterion to check if our Cauchy problem is non-characteristic. For an initial curve $\gamma$ and a PDE a $u_x+bu_y=c$ it is given by 
$$\begin{vmatrix}a|_\gamma & x_0' \\ b|_\gamma & y_0'\end{vmatrix}\neq0 \Longrightarrow \text{non-characteristic Cauchy problem} $$
(a) $~\begin{vmatrix} 1 & 0 \\ y_0 & 1\end{vmatrix}=1 \neq 0$
$$\begin{cases} x_t&=\cos(x)^2 \\ y_t&=y \\ u_t&=0 \end{cases} \Rightarrow \begin{cases} x&=\arctan(t) \\ y&=y_0 e^{t}\\ u&=u_0=y_0^2 \end{cases} \Rightarrow t=\tan(x), y=y_0 \exp(\tan(x))$$ $$\Longrightarrow u(x,y)=\frac{y^2}{\exp(2\tan(x))}$$
You can easily check that this is indeed the solution to this Cauchy problem.
(b) $~\begin{vmatrix} \cos^2(x_0) & 1 \\ 0 & 0\end{vmatrix}=0$ 
$$\begin{cases} x_t&=\cos(x)^2 \\ y_t&=y \\ u_t&=0 \end{cases} \Rightarrow  \begin{cases} x&=\arctan(t+\tan(x_0)) \\ y&=0 \\ u&=x_0^2 \end{cases}$$
As you noticed correctly this system is underdetermined.
