There are other substitutions you can use to transform the equation to a normal form. For example, the equation
$$
a(x)y''(x)+b(x)y'(x)+c(x)y(x)+\lambda d(x)y(x)=0,
$$
is transformed to
$$
-\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f = \lambda d(x)f(x)
$$
by the substitution $y=\rho(x)f(x)$, where
$$
\rho(x) = \sqrt{a(x)}\exp\left(-\int\frac{b}{2a}dx\right).
$$
This substitution is far more versatile in dealing with classical equations. However, it requires $a(x)$ to be twice differentiable. The advantage is that $a(x)$ remains the coefficient of $f''$ and $d(x)$ remains the of the eigenvalue $\lambda f(x)$, which is what I believe you're after. Many classical standard forms can be derived using this substitution.
For example, the associated Legendre equation
$$
(1-x^2)u''-2x(m+1)u'+[l(l+1)-m^2-m]u=0
$$
is transformed by $u=(1-x^2)^{-m/2}f$ to the classical form
$$
-\frac{d}{dx}\left((1-x^2)\frac{df}{d}\right)+\frac{m^2}{1-x^2}f=l(l+1)f.
$$
This form is hard to get any other way.
Another use of this substitution is to transform an equation to potential form
$$
-\frac{d^2f}{dx^2}+qf = \lambda f.
$$
This requires a clever use of the above substitution, followed by a substitution of the independent variable.
ANOTHER APPROACH: Typically, the function $d$ is turned into a weight function for the inner product space when you have a form such as the one I mentioned above:
$$
-\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f = \lambda d(x)f(x)
$$
Then you can define the operator
$$
L=\frac{1}{d}\left[ -\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f \right]
$$
on the weighted $L^2$ space $L^2_{d}$ defined by
$$
\langle f,g\rangle_{L^2_{d}}=\int f(x)\overline{g(x)}d(x)dx
$$
In this setting the operator $L$ is symmetric with the right endpoint conditions:
$$
\langle Lf,g\rangle_{L^2_d}=\langle f,Lg\rangle_{L^2_d}+\mbox{evaluation terms},
$$
and the eigenvalue problem is $Lf=\lambda f$.
Following up on my comment, given
$$y'' + 2y' + (\lambda + 1)y = 0$$
You can multiply by the "integrating factor" $\mu(x) = e^{\int 2 dx} = e^{2x}$,
$$e^{2x}y'' + 2e^{2x}y' + (\lambda + 1)e^{2x}y = 0$$
At which point you have
$$ \frac{d}{dx}\left(e^{2x} y' \right) + (e^{2x} + \lambda e^{2x})y = 0 $$
as desired.
Best Answer
That's correct. We just multiply by $e^{-x}$: $$ \underbrace{e^{-x}xy'' + e^{-x}(1 - x)y'}_{=\left(xe^{-x}y'\right)'} + λe^{-x}y = 0, $$ and then we get $$ \left(xe^{-x}y'\right)'+λe^{-x}y = 0. $$