The problem with differentiating the PDEs is that you don't know a priori how differentiable the functions are i.e whether the functions are $C^{1}, C^{2}$ etc. If you assume the solutions are regular enough, then there is no problem (and if this is for an undergraduate class, this is probably assumed). However, you can avoid these regularity issues by using the method of characteristics. Your problem in matrix form is
\begin{align}
\begin{pmatrix}
h \\
u
\end{pmatrix}_{t} &= \begin{pmatrix}
0 & - H_{0} \\
- g & 0
\end{pmatrix}
\begin{pmatrix}
h \\
u
\end{pmatrix}_{x} \\
\implies \vec{h}_{t} &= M \vec{h}_{x}
\end{align}
Computing the eigenvalues and eigenvectors of $M$ allows us to diagonalise $M$ as $M = P D P^{-1}$, where
\begin{align}
P &= \begin{pmatrix}
\sqrt{H_{0}/g} & -\sqrt{H_{0}/g} \\
1 & 1
\end{pmatrix} \\\\
\implies P^{-1} &= \frac{1}{2 \sqrt{H_{0}/g}} \begin{pmatrix}
1 & \sqrt{H_{0}/g} \\
-1 & \sqrt{H_{0}/g}
\end{pmatrix} \\\\
D &= \begin{pmatrix}
\sqrt{g H_{0}} & 0 \\
0 & -\sqrt{g H_{0}}
\end{pmatrix}
\end{align}
and hence your equation set becomes
$$\vec{h}_{t} = P D P^{-1} \vec{h}_{x}$$
Multiplying both sides by $P^{-1}$ and setting $P^{-1} \vec{h} = \vec{v} = (v_{1}, v_{2})^{T}$ yields the system
$$\vec{v}_{t} = D \vec{v}_{x}$$
which is directly solvable
\begin{align}
v_{1} &= f_{1}(x + \sqrt{g H_{0}} t) \\
v_{2} &= f_{2}(x - \sqrt{g H_{0}} t)
\end{align}
with $f_{1}, f_{2}$ arbitrary differentiable functions. Inverting the relationship $P^{-1} \vec{h} = \vec{v} \implies \vec{h} = P \vec{v}$ yields the general solution to the original problem
\begin{align}
h &= \sqrt{H_{0}/g} v_{1} - \sqrt{H_{0}/g} v_{2} \\
u &= v_{1} + v_{2}
\end{align}
after which the initial conditions can be applied.
$$(1-x^2)\cdot y''_{xx} -x\cdot y'_x +y = \frac{\sqrt{1-x^2}}{x}$$
I don't know about the given hint but you can rewrite the DE as:
$$((1-x^2)\cdot y')'+(xy)' = \frac{\sqrt{1-x^2}}{x}$$
Integrate.
Note that we have:
$$x\dfrac {dy}{dx}=\cos t \dfrac {dy}{dt}\dfrac {dt}{dx}=-y'_t\dfrac {\cos t}{\sin t}$$
so that you have $$\dfrac {dy}{dx}=-\dfrac {dy}{dt}\dfrac 1 {\sin t}$$
Then you calculate $(1-x^2)y''_{xx}$
You should get:
$$y''_{tt}-y'_{t}\dfrac {\cos t }{\sin t}$$
This simplify your differential equation.
$$y''_{tt}+y=\tan t$$
It seems to me that you made a mistake when you calculated $y''_{tt}$.
Best Answer
I have some difficulty to understand what you did...
Anyway, I think that separation of variables is the easiest way to solve it (this should be the same thing you tried, I guess :D): $$y' = (y + 1)x$$ $$\frac{y'}{y + 1} = x$$ and formally, $$\frac{dy}{y+1} = x\ dx.$$ Integrating you obtain $$\ln(1+y) = \frac{x^2}{2} + c,$$ therefore, $$ 1 + y = e^{\ln(1+y)} = e^{\frac{x^2}{2}}e^{c}.$$ This gives $$y(x) = e^{\frac{x^2}{2}}e^c -1.$$You need to use the Cauchy's condition to find the solution of the equation (if $x = 1$ then $y = 2$): $$2 = e^{\frac{1}{2}} \cdot e^c - 1 \Longrightarrow e^c = 3e^{-\frac{1}{2}}.$$ The only solution of your Cauchy problem is $$y(x) = 3 e^{\frac{x^2}{2} - \frac{1}{2}} -1.$$
Note that the passage in which we divided by $1 + y$ is justified by the fact that we are solving in a neighborhood of $x_0 = 1$, where $1 + y \neq 0$. This is possible being $y$ countinuous.
I hope this can help!!!