It's a linear homogeneous equation, you can lower the order of equation by the following, general method:

\(\displaystyle{y}{\left({x}\right)}={e}^{{{s}{\left({x}\right)}}}\)

\(\displaystyle{y}'={s}'{e}^{{s}}\)

\(\displaystyle{y}{''}={s}{''}{e}^{{s}}+{s}'^{{2}}{e}^{{s}}\)

Substituting this and dividing by \(\displaystyle{e}^{{s}}\), you have:

\(\displaystyle{\left({x}+{1}\right)}{s}{''}+{\left({x}+{1}\right)}{s}'^{{2}}-{\left({2}{x}+{3}\right)}{s}'+{\left({x}+{2}\right)}={0}\)

Now set:

\(\displaystyle{s}'{\left({x}\right)}={p}{\left({x}\right)}\)

You obtain a first order nonlinear ode:

\(\displaystyle{\left({x}+{1}\right)}{p}'+{\left({x}+{1}\right)}{p}^{{2}}-{\left({2}{x}+{3}\right)}{p}+{\left({x}+{2}\right)}={0}\)

It's a Riccati equation

The nice thing about Riccati equation is that if you know any particular solution, there's a way to express the general solution through it.

I'll expand more on this below:

Let \(\displaystyle{p}_{{0}}\) be a particular solution to (*). In the OP case it's:

\(\displaystyle{p}_{{0}}={1}\)

Now let the general solution be still denoted as p. Substitute the sum of these two solutions in (*):

\(\displaystyle{\left({x}+{1}\right)}{\left({p}_{{0}}'+{p}'\right)}+{\left({x}+{1}\right)}{\left({{p}_{{0}}^{{2}}}+{2}{p}_{{0}}{p}+{p}^{{2}}\right)}-{\left({2}{x}+{3}\right)}{\left({p}_{{0}}+{p}\right)}+{\left({x}+{2}\right)}={0}\)

Note that because (*) is nonlinear the sum of two solutions is not a solution to the original equation anymore. However, since each of them separately is still a solution, we can substitute

and see:
\(\displaystyle{p}_{{0}}={1}\)

\(\displaystyle{\left({x}+{1}\right)}{p}'+{\left({x}+{1}\right)}{\left({1}+{2}{p}+{p}^{{2}}\right)}-{\left({2}{x}+{3}\right)}{\left({1}+{p}\right)}+{\left({x}+{2}\right)}={0}\)

Or simplifying:

\(\displaystyle{\left({x}+{1}\right)}{p}'+{\left({x}+{1}\right)}{p}^{{2}}-{p}={0}\)

The equation (**) is a Bernoulli equation and can be solved by the following method:

\(\displaystyle{q}{\left({x}\right)}={\frac{{{1}}}{{{p}{\left({x}\right)}}}}\)

\(\displaystyle-{\left({x}+{1}\right)}{\frac{{{q}'}}{{{q}^{{2}}}}}+{\left({x}+{1}\right)}{\frac{{{1}}}{{{q}^{{2}}}}}-{\frac{{{1}}}{{{q}}}}={0}\)

\(\displaystyle{q}'-{1}+{\frac{{{q}}}{{{\left({1}+{x}\right)}}}}={0}\)

Now (3) is finally a linear first odrer ODE as promised. Going back through all the substitutions will bring the general solution of the original equation