First term for particular solution of series solution of inhomogenous DE Watch

Baal Hadad
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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.
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Y12_FurtherMaths
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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?
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Baal Hadad
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(Original post by Y12_FurtherMaths)
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?
OK...
The complete solution ,y=c0(y1)+c1(y2)+yp
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Baal Hadad
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(Original post by Y12_FurtherMaths)
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?
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Y12_FurtherMaths
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(Original post by Baal Hadad)
Is it OK? Or you need additional information?
I'm afraid I'm unable to help you with this. Is this university maths?
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Baal Hadad
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(Original post by Y12_FurtherMaths)
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.
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Y12_FurtherMaths
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(Original post by Baal Hadad)
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
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Baal Hadad
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(Original post by Y12_FurtherMaths)
It doesn't seem like people will be able to help you im afraid
Never mind,nobody is obligated to help me here.Thanks.By the way,in my previous reply,I mean that I am not entering uni yet.
Last edited by Baal Hadad; 4 weeks ago
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RDKGames
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(Original post by Baal Hadad)
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.
Last edited by RDKGames; 4 weeks ago
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Baal Hadad
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(Original post by RDKGames)
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.
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?
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Baal Hadad
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(Original post by RDKGames)
b_1 = -1 .... (from comparing coefficients of x^{-1})
And forgive me,is this x^{1} instead of x^{-1} ?
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RDKGames
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(Original post by Baal Hadad)
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?
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.

(Original post by Baal Hadad)
And forgive me,is this x^{1} instead of x^{-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.
Last edited by RDKGames; 4 weeks ago
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Baal Hadad
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(Original post by RDKGames)
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.
Yes,just like what you had said,during my calculations I found that b1 is determined and b2 depends on b0.OK,after your explainations I accepted the choice of choosing zero.

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?
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RDKGames
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(Original post by Baal Hadad)
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.
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Baal Hadad
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(Original post by RDKGames)
Yep.
Your explanation helps me a lot,since I can't ask my friends and teachers (I am graduated from high school)Thanks for your help!
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