# The differential equation is as follows: (x+1)y''-(2x+3)y'+(x+2)y=0

The differential equation is as follows: $$\displaystyle{\left({x}+{1}\right)}{y}{''}-{\left({2}{x}+{3}\right)}{y}'+{\left({x}+{2}\right)}{y}={0}$$ where y(x) is a funtion on $$\displaystyle{\mathbb{{{R}}}}$$. First, I had to show that $$\displaystyle{y}_{{1}}{\left({x}\right)}={e}^{{x}}$$ is a solution for the DV. The second question was to find the general solution of the DV. I instantly started on the series of solution solving method, but I think I have to use the 'hint' $$\displaystyle{y}_{{1}}{\left({x}\right)}={e}^{{x}}$$ is a solution to the DV, but how to do this?

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Alrew1959
Let
$$\displaystyle{y}{\left({x}\right)}={y}_{{1}}{\left({x}\right)}{z}{\left({x}\right)}={e}^{{x}}{z}{\left({x}\right)}$$
where z(x) is your new unknown function. Then $$\displaystyle{y}'{\left({x}\right)}={e}^{{x}}{z}{\left({x}\right)}+{e}^{{x}}{z}'{\left({x}\right)}$$ and $$\displaystyle{y}{''}{\left({x}\right)}=..$$, and when you substitute this into your ODE you get something with a z'' term and a z' term, but no z term, so if you let $$\displaystyle{w}{\left({x}\right)}={z}'{\left({x}\right)}$$ you get a first-order linear ODE for $$\displaystyle{w}$$, and such equations can always be solved in principle, using an integrating factor.
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Uersfeldte
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