First term for particular solution of series solution of inhomogenous DE

I doubt many people are going to read this because it looks absolutely horrendous typed out! Can you perhaps upload a picture of it instead?

I doubt many people are going to read this because it looks absolutely horrendous typed out! Can you perhaps upload a picture of it instead?

Is it OK? Or you need additional information?

I'm afraid I'm unable to help you with this. Is this university maths?

I don't know,I am self studying for this and I still don't have a uni degree.But I think series solution is not covered in A Level syllabus.

It doesn't seem like people will be able to help you im afraid

I am solving a differential equation:
x*y''(x)+(x+1)*y'(x)+x*y(x) = x^2+1
Firstly,I tried to find the complementary solution.
Using Frobenius' method and expanding around x=0,
the first solution is
1-(1/4)*x^2+(1/18)*x^3+(1/192)*x^4-(11/3600)*x^5+(29/103680)*x^6+O(x^7)
the second solution is
ln(x)*(1-(1/4)*x^2+(1/18)*x^3+(1/192)*x^4-(11/3600)*x^5+(29/103680)*x^6+O(x^7))-x+(1/2)*x^2-(1/108)*x^3-(41/1152)*x^4+(1529/216000)*x^5-(1/345600)*x^6+O(x^7)
Since the indicial equation has repeated solutions,both equals to zero.
After that,I tried to find the particular solution.
I assumed the particular solution has the form of a series,
y=c0+c1(x)+c2(x^2)+O(x^3)
and tried to compare the coefficients.
Although I obtained the particular solution
x*(1-(1/4)*x+(1/18)*x^2+(1/192)*x^3-(11/3600)*x^4+(29/103680)*x^5+O(x^6))
,which is correct after checking using Maple,but I assumed c0=0 to obtain this solution,but why?
And also,when finding P(x) in y2=y1(ln x)+P(x)(after substituting the solution of indicial equation)
,I also had to assume that the term without the unknown is 0,is it because the indicial equation has repeated roots?
Sorry for my bad English and my ignorance since I am self-studying in this topic.
By the way,I am self-studying based on this book:
(Undergraduate Lecture Notes in Physics) Lev Kantorovich - Mathematics for Natural Scientists_ Fundamentals and Basics-Springer (2015)
Thanks.

This is university maths. It goes deep into the theory of Linear Differential Equations.

So to answer your question, I'll run you down the problem quickly.

Firstly, you put the ODE into canonical form

$y'' + \dfrac{x+1}{x}y' + y = \dfrac{x^2+1}{x}$.


It is clear that $x=0$ is a regular singular point, therefore Method of Frobenius is the way to go.

Solving the homogeneous case, you found the solution

$y_1 = \displaystyle \sum_{n=0}^{\infty} a_n x^n = 1-\dfrac{1}{4}x^2+\dfrac{1}{18}x^3 + \dfrac{1}{192}x^4 - \dfrac{11}{3600}x^5+O(x^6)$.

The second solution, because the roots repeat, is

$y_2 = y_1\ln x + \displaystyle \sum_{n=0}^{\infty} b_nx^n$.

If you substitute this into the homogeneous case and compare coefficients, the first two comparisons would yield you

$b_1 = -1$ .... (from comparing coefficients of $x^{-1}$)

and

$4b_2 - 2 + b_0 = 0$ (*) .... (from comparing coefficients of $x^0$)

Any further comparisons will not isolate $b_0$ for you, in fact none of them will add on any extra restrictions on it. So in fact, you can choose it arbitrarily as long as equation (*) holds. Of course, whenever we say arbitrarily, we usually mean try to choose something that makes the problem simple. So of course, we can just choose $b_0 = 0$. This implies $b_2 = \dfrac{1}{2}$, and so all of our $b_n$ are now fully defined.


Same approach applies to when you try a particular solution

$y_P = \displaystyle \sum_{n=0}^{\infty} c_n x^n$.

You run into the issue of having no restriction on $c_0$ but you want your $y_P$ to be defined, so just choose it to be zero.

Hence you obtain the solution that Maple is throwing at you

$y(x) = Ay_1 + By_2 + y_P$.

$b_1 = -1$ .... (from comparing coefficients of $x^{-1}$)

Well,do you mean that both the terms can be arbitrarily chosen,so 0 is chosen to make the calculation more easier? If we choose another number other than zero,does the resulted series is just the series using 0 plus multiple of y1?

And forgive me,is this $x^{1}$ instead of $x^{-1}$ ?

In the series $\displaystyle \sum_{n=0}^{\infty} b_nx^n$ you find that $b_1 = -1$ but $b_2$ ultimately depends on what $b_0$ is. In fact, every $b_n$ for $n \geq 2$ depends on what $b_0$ is. Once you set that, every other coefficient gets fixed in place.

This coefficient is an arbitrary choice for us, because we have no restriction for it. Any suitable value of it will yield every other $b_n$ value and hence give us a solution for $y_2$. If that makes sense to you, then the next step for you to understand is that $b_0 = 0$ is the simplest case for us pick, so just do that.

If you pick a non-zero value then you're going to get the same sum but expressed differently.



You do the same thing when you determine the solution for $y_1 = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. You had the choice of picking arbitrary $a_0$ but you picked $a_0 = 1$ since the obtained recurrence relation must satisfy

$a_n = \dfrac{(1-n)a_{n-1} - a_{n-2}}{n^2}$

so $a_1 = 0$ (from comparing $x^{-1}$ terms) and $a_2 = -\dfrac{a_0}{4}$. Similarly, $a_3$ depends on $a_0$, and so on. You fixed $a_0 = 1$ and obtained all the coefficients you see in $y_1$.



No. Once you substitute in a series into this particular ODE, you should notice that we always begin with $x^{-1}$, then we have the $x^0$ (constant) terms, and then $x^1$, and so on... but for my explanation I'm only considering $x^{-1}$ and $x^0$.

The $x^{-1}$ coefficients turn out to be $a_1$ (when you substitute in $y_1$) and $b_1 + 1$ (when you substitute in $y_2$) which yields us $a_1 = 0$ and $b_1 = -1$. But once again I stress the fact that we have no restrictions on $a_0, b_0$ hence we choose them arbitrarily so we just pick $a_0 = 1$ and $b_0 = 0$.

I say they're arbitrary, for you shouldn't (for instance) pick $a_0 = 0$ because then the entire series for $y_1$ is just zero -- a trivial solution.

Well,I think you use x^(-1) and I use x^(0) because you substituted the series to the canonical form of the equation but I substituted it to the original equation in the first post.Am I right?

Yep.