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

folklorahhe 2021-11-19 Answered
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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Expert Answer

Alrew1959
Answered 2021-11-20 Author has 774 answers
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
Answered 2021-11-21 Author has 1251 answers
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
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