I am trying to solve the following linear pde where $u=f(x,y)$ in the domain $y \in  (0,\infty)$: 
$$y\dfrac{\partial{u}}{\partial x} = \dfrac{\partial^2 u}{\partial y^2}$$
with boundary conditions: 
$$u(x,0)=\sin(x) $$ 
$$\lim_{y \rightarrow \infty} u(x,y) = 0 .$$

Can someone please suggest how do I proceed to get the analytical solution for this equation? 



Solution: 

Using method of separation of variables, 
we assume solution to be of the form: $$u(x,y)=X(x).Y(y) $$
Inserting this to the pde, we get: 
$$yX'Y=XY'' $$
$$\dfrac{X'}{X}=\dfrac{Y''}{yY}=-\lambda$$
Now we get the following ode's 
$$X'+\lambda X=0 $$
$$Y'' + \lambda y Y = 0 $$

from which we get two general solutions of the following form: 
$$X(x) = c_1\exp(-\lambda x) \qquad Y(y) = c_2Ai(\lambda^{1/3} y) + c_3Bi(\lambda^{1/3} y)  $$ where $Ai$ and $Bi $ are Airy's functions and the general solution can be expressed as: 
$$u(x,y) =  c_1\exp(-\lambda x).(c_2Ai(\lambda^{1/3} y) + c_3Bi(\lambda^{1/3} y))$$

$c_3 \rightarrow 0 $ to satisfy second boundary condition since $Bi(\infty) \rightarrow \infty $ and hence, solution is of the form: 

$$u(x,y) = A \exp(-\lambda x) Ai(\lambda^{1/3} y) $$
Putting $y=0$, 
$$u(x,0) = A\exp(-\lambda x).\dfrac{1}{3^{2/3}}\Gamma({2/3}) = \sin(x) $$

How should I now proceed to find $A$ and $\lambda$, considering that $\lambda$ needs to be positive real value?


Question: Solution disappearance responsive navigation after open/close
Answer: You might want to add a new media query like i did here @ FIDDLE

@media all and (min-width: 1000px) {
  nav ul { 
    display: block;
    top: 60px;
  }
}