Using $y=uv$ substitution to solve $y'' + P(x)y' + Q(x)y = 0$ 
Show that by applying the substitution $y=uv$ to the homogeneous
  equation $y'' + P(x)y' + Q(x)y = 0$, it is possible to obtain a
  homogeneous second order linear equation for $v$ with no $v′$ term
  present. Find $u$ and the equation for $v$ in terms of the original
  coefficients $P(x)$ and $Q(x)$

This problem is given in GF Simmons book on differential equations Pg $119$ Problem $11$, $3rd$ edition.
I am confused. Here $u$ and $v$ are supposed to be dependent on $x$ so how am I supposed to differentiate these and replace derivatives of $y$ and get only a differential equation in $v$?

Further I am also asked to solve $y″+2xy′+(1 + x^2)y=0$ using this
  method. I am hoping I would be able to solve this if someone could
  explain me what we're trying to do in this method

 A: No.  $u$ and $v$ are functions of $x$.  You are factoring $y$ as a product of two functions.  There are a lot of such pairs.  Then you are free to move through the space of such pairs until you find a pair that happens to make all the $v'$s cancel in the equation.
Making the substitution $y(x) = u(x) v(x)$, we obtain the equation 
$$  \hspace{-0.5in}\underbrace{u v'' + 2u'v' + v u''}_{y''} + P(x)(\underbrace{u'v + uv'}_{y'}) + Q(x)\underbrace{uv}_{y} = 0  \tag*{(*)}$$
So all we need to find is a pair of $u$ and $v$ such that $y = uv$ and 
$$  2u'v' + P(x)uv' = 0  \text{.}  \tag{**}$$
This factors, so either $v' = 0$ (meaning $v$ is just a constant, which is unlikely to be a useful factorization of $y$) or $2u'+P(x)u = 0$.  We can solve this first order ODE:
$$  u(x) = \exp \left(C + \int_1^x \; -\frac{1}{2} P(t)\,\mathrm{d}t  \right)  \text{,}  $$
where $C$ is a constant of integration.
With this choice of $u$, (**) is satisfied for any choice of $v(x)$, so let $v(x) = y(x)/u(x)$.  Then eliminating $v'$ between (*) and (**), we have
$$  u v'' + v u'' + P(x)u'v + Q(x)uv = 0  $$
or, what is perhaps more clearly what is requested,
$$  u v'' + \left(u'' + P(x)u' + Q(x)u \right)v = 0  \text{.}  $$
A: $$y=uv \implies y'=u'v+uv' $$
$$\implies y''=u''v+2u'v'+uv''$$
Back to your equation:
$$y″+2xy′+(1 + x^2)y=0$$
$$u''v+2u'v'+uv''+2x(u'v+uv')+(1 + x^2)uv=0$$
$$u''v+2u'(v'+xv)+u(v''+2xv'+(1 + x^2)v)=0$$
Then suppose 
$$v'+xv=0 \implies v''+v+xv'=0$$
The equation becomes:
$$u''v+xv'+ x^2v=0$$
$$u''=0$$
Now you have to solve this:
$$
\begin{cases}
u''=0 \\
v'+xv=0
\end{cases}
$$
$$
\begin{cases}
u=c_1x+c_2\\
v=e^{-x^2/2} \\
\end{cases}
$$
$$\implies y(x)=uv=(c_1x+c_2)e^{-x^2/2} $$
