Solve ODE using Fourier series.












2












$begingroup$



I have that



$$f(x)=frac{4}{3}+frac{2}{pi^2}sum_{ninmathbb{Z}}frac{(-1)^n(1+ipi
n)}{n^2}e^{ipi n x}, quad nneq0tag1$$



and I want to find a 2-periodic function $y(x)$ that solves the
following diff-equation



$$2y''-y'-y=f(x).tag{2}$$




Since $(1)$ is a linear ODE I know that it's solution is of the form $y(x)=y_h+y_p.$ Finding $y_h$ is trivial and done by solving the characteristic equation. I found it to be $y_h=C_1e^x+C_2e^{-x/2}.$



For $y_p$, I need to use $f(x)$ somehow, I could not find any good example in the book on how to do this. I only found one video on youtube but it does not really seem to be that similar to my problem.



I want to follow the answer in this thread but I can't seem to grasp the mechanics of what he means. Can someone show me how to go about this?






ordinary-differential-equations fourier-analysis fourier-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
    $endgroup$
    – Dylan
    Jan 24 at 12:02












  • $begingroup$
    @Dylan - How do I write the solution in series form if I don't know the solution yet?
    $endgroup$
    – Parseval
    Jan 24 at 12:34










  • $begingroup$
    You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
    $endgroup$
    – Dylan
    Jan 24 at 13:00


















2












$begingroup$



I have that



$$f(x)=frac{4}{3}+frac{2}{pi^2}sum_{ninmathbb{Z}}frac{(-1)^n(1+ipi
n)}{n^2}e^{ipi n x}, quad nneq0tag1$$



and I want to find a 2-periodic function $y(x)$ that solves the
following diff-equation



$$2y''-y'-y=f(x).tag{2}$$




Since $(1)$ is a linear ODE I know that it's solution is of the form $y(x)=y_h+y_p.$ Finding $y_h$ is trivial and done by solving the characteristic equation. I found it to be $y_h=C_1e^x+C_2e^{-x/2}.$



For $y_p$, I need to use $f(x)$ somehow, I could not find any good example in the book on how to do this. I only found one video on youtube but it does not really seem to be that similar to my problem.



I want to follow the answer in this thread but I can't seem to grasp the mechanics of what he means. Can someone show me how to go about this?






ordinary-differential-equations fourier-analysis fourier-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
    $endgroup$
    – Dylan
    Jan 24 at 12:02












  • $begingroup$
    @Dylan - How do I write the solution in series form if I don't know the solution yet?
    $endgroup$
    – Parseval
    Jan 24 at 12:34










  • $begingroup$
    You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
    $endgroup$
    – Dylan
    Jan 24 at 13:00
















2












2








2





$begingroup$



I have that



$$f(x)=frac{4}{3}+frac{2}{pi^2}sum_{ninmathbb{Z}}frac{(-1)^n(1+ipi
n)}{n^2}e^{ipi n x}, quad nneq0tag1$$



and I want to find a 2-periodic function $y(x)$ that solves the
following diff-equation



$$2y''-y'-y=f(x).tag{2}$$




Since $(1)$ is a linear ODE I know that it's solution is of the form $y(x)=y_h+y_p.$ Finding $y_h$ is trivial and done by solving the characteristic equation. I found it to be $y_h=C_1e^x+C_2e^{-x/2}.$



For $y_p$, I need to use $f(x)$ somehow, I could not find any good example in the book on how to do this. I only found one video on youtube but it does not really seem to be that similar to my problem.



I want to follow the answer in this thread but I can't seem to grasp the mechanics of what he means. Can someone show me how to go about this?






ordinary-differential-equations fourier-analysis fourier-series







share|cite|improve this question











$endgroup$





I have that



$$f(x)=frac{4}{3}+frac{2}{pi^2}sum_{ninmathbb{Z}}frac{(-1)^n(1+ipi
n)}{n^2}e^{ipi n x}, quad nneq0tag1$$



and I want to find a 2-periodic function $y(x)$ that solves the
following diff-equation



$$2y''-y'-y=f(x).tag{2}$$




Since $(1)$ is a linear ODE I know that it's solution is of the form $y(x)=y_h+y_p.$ Finding $y_h$ is trivial and done by solving the characteristic equation. I found it to be $y_h=C_1e^x+C_2e^{-x/2}.$



For $y_p$, I need to use $f(x)$ somehow, I could not find any good example in the book on how to do this. I only found one video on youtube but it does not really seem to be that similar to my problem.



I want to follow the answer in this thread but I can't seem to grasp the mechanics of what he means. Can someone show me how to go about this?






ordinary-differential-equations fourier-analysis fourier-series




ordinary-differential-equations fourier-analysis fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 12:08









Dylan

13.6k31027




13.6k31027










asked Jan 24 at 11:55









ParsevalParseval

2,9881719




2,9881719








  • 1




    $begingroup$
    Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
    $endgroup$
    – Dylan
    Jan 24 at 12:02












  • $begingroup$
    @Dylan - How do I write the solution in series form if I don't know the solution yet?
    $endgroup$
    – Parseval
    Jan 24 at 12:34










  • $begingroup$
    You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
    $endgroup$
    – Dylan
    Jan 24 at 13:00
















  • 1




    $begingroup$
    Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
    $endgroup$
    – Dylan
    Jan 24 at 12:02












  • $begingroup$
    @Dylan - How do I write the solution in series form if I don't know the solution yet?
    $endgroup$
    – Parseval
    Jan 24 at 12:34










  • $begingroup$
    You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
    $endgroup$
    – Dylan
    Jan 24 at 13:00










1




1




$begingroup$
Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
$endgroup$
– Dylan
Jan 24 at 12:02






$begingroup$
Write the solution in its series form and plug it into the ODE to obtain a system of algebraic equations in terms of the Fourier coefficients. You already have $f$ in series form so this should be fairly straightforward
$endgroup$
– Dylan
Jan 24 at 12:02














$begingroup$
@Dylan - How do I write the solution in series form if I don't know the solution yet?
$endgroup$
– Parseval
Jan 24 at 12:34




$begingroup$
@Dylan - How do I write the solution in series form if I don't know the solution yet?
$endgroup$
– Parseval
Jan 24 at 12:34












$begingroup$
You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
$endgroup$
– Dylan
Jan 24 at 13:00






$begingroup$
You write it as an unknown series, and derive the coefficients from the resulting equations. See my answer.
$endgroup$
– Dylan
Jan 24 at 13:00












1 Answer
1






active

oldest

votes


















1












$begingroup$

A $2$-periodic function has the form



$$ y(x) = c_0 + sum_{n=pm 1}^{pminfty} c_n e^{inpi x} $$



where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)



$$ 2y'' - y' - y = - c_0 - sum_{n=pm1}^{pminfty} (n^2pi^2 + inpi + 1)c_n e^{inpi x} $$



Comparing coefficients gives



begin{align}
-c_0 &= frac43 \
-(n^2pi^2+inpi+1)c_n &= frac{2}{pi^2}(-1)^n frac{1+inpi}{n^2}
end{align}



The homogeneous solution is not periodic so it isn't needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
    $endgroup$
    – Parseval
    Jan 24 at 14:32










  • $begingroup$
    $2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
    $endgroup$
    – Dylan
    Jan 24 at 14:45










  • $begingroup$
    Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
    $endgroup$
    – Parseval
    Jan 24 at 14:56






  • 1




    $begingroup$
    The first series does include a constant term. What is $e^{inx}$ when $n=0$?
    $endgroup$
    – Dylan
    Jan 24 at 15:03










  • $begingroup$
    Ok I got it now! Thanks.
    $endgroup$
    – Parseval
    Jan 24 at 15:08











Your Answer





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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

A $2$-periodic function has the form



$$ y(x) = c_0 + sum_{n=pm 1}^{pminfty} c_n e^{inpi x} $$



where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)



$$ 2y'' - y' - y = - c_0 - sum_{n=pm1}^{pminfty} (n^2pi^2 + inpi + 1)c_n e^{inpi x} $$



Comparing coefficients gives



begin{align}
-c_0 &= frac43 \
-(n^2pi^2+inpi+1)c_n &= frac{2}{pi^2}(-1)^n frac{1+inpi}{n^2}
end{align}



The homogeneous solution is not periodic so it isn't needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
    $endgroup$
    – Parseval
    Jan 24 at 14:32










  • $begingroup$
    $2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
    $endgroup$
    – Dylan
    Jan 24 at 14:45










  • $begingroup$
    Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
    $endgroup$
    – Parseval
    Jan 24 at 14:56






  • 1




    $begingroup$
    The first series does include a constant term. What is $e^{inx}$ when $n=0$?
    $endgroup$
    – Dylan
    Jan 24 at 15:03










  • $begingroup$
    Ok I got it now! Thanks.
    $endgroup$
    – Parseval
    Jan 24 at 15:08
















1












$begingroup$

A $2$-periodic function has the form



$$ y(x) = c_0 + sum_{n=pm 1}^{pminfty} c_n e^{inpi x} $$



where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)



$$ 2y'' - y' - y = - c_0 - sum_{n=pm1}^{pminfty} (n^2pi^2 + inpi + 1)c_n e^{inpi x} $$



Comparing coefficients gives



begin{align}
-c_0 &= frac43 \
-(n^2pi^2+inpi+1)c_n &= frac{2}{pi^2}(-1)^n frac{1+inpi}{n^2}
end{align}



The homogeneous solution is not periodic so it isn't needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
    $endgroup$
    – Parseval
    Jan 24 at 14:32










  • $begingroup$
    $2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
    $endgroup$
    – Dylan
    Jan 24 at 14:45










  • $begingroup$
    Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
    $endgroup$
    – Parseval
    Jan 24 at 14:56






  • 1




    $begingroup$
    The first series does include a constant term. What is $e^{inx}$ when $n=0$?
    $endgroup$
    – Dylan
    Jan 24 at 15:03










  • $begingroup$
    Ok I got it now! Thanks.
    $endgroup$
    – Parseval
    Jan 24 at 15:08














1












1








1





$begingroup$

A $2$-periodic function has the form



$$ y(x) = c_0 + sum_{n=pm 1}^{pminfty} c_n e^{inpi x} $$



where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)



$$ 2y'' - y' - y = - c_0 - sum_{n=pm1}^{pminfty} (n^2pi^2 + inpi + 1)c_n e^{inpi x} $$



Comparing coefficients gives



begin{align}
-c_0 &= frac43 \
-(n^2pi^2+inpi+1)c_n &= frac{2}{pi^2}(-1)^n frac{1+inpi}{n^2}
end{align}



The homogeneous solution is not periodic so it isn't needed.






share|cite|improve this answer









$endgroup$



A $2$-periodic function has the form



$$ y(x) = c_0 + sum_{n=pm 1}^{pminfty} c_n e^{inpi x} $$



where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)



$$ 2y'' - y' - y = - c_0 - sum_{n=pm1}^{pminfty} (n^2pi^2 + inpi + 1)c_n e^{inpi x} $$



Comparing coefficients gives



begin{align}
-c_0 &= frac43 \
-(n^2pi^2+inpi+1)c_n &= frac{2}{pi^2}(-1)^n frac{1+inpi}{n^2}
end{align}



The homogeneous solution is not periodic so it isn't needed.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 12:54









DylanDylan

13.6k31027




13.6k31027












  • $begingroup$
    This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
    $endgroup$
    – Parseval
    Jan 24 at 14:32










  • $begingroup$
    $2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
    $endgroup$
    – Dylan
    Jan 24 at 14:45










  • $begingroup$
    Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
    $endgroup$
    – Parseval
    Jan 24 at 14:56






  • 1




    $begingroup$
    The first series does include a constant term. What is $e^{inx}$ when $n=0$?
    $endgroup$
    – Dylan
    Jan 24 at 15:03










  • $begingroup$
    Ok I got it now! Thanks.
    $endgroup$
    – Parseval
    Jan 24 at 15:08


















  • $begingroup$
    This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
    $endgroup$
    – Parseval
    Jan 24 at 14:32










  • $begingroup$
    $2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
    $endgroup$
    – Dylan
    Jan 24 at 14:45










  • $begingroup$
    Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
    $endgroup$
    – Parseval
    Jan 24 at 14:56






  • 1




    $begingroup$
    The first series does include a constant term. What is $e^{inx}$ when $n=0$?
    $endgroup$
    – Dylan
    Jan 24 at 15:03










  • $begingroup$
    Ok I got it now! Thanks.
    $endgroup$
    – Parseval
    Jan 24 at 15:08
















$begingroup$
This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
$endgroup$
– Parseval
Jan 24 at 14:32




$begingroup$
This is great, thank you! I have a question, what is meant by "2-periodic"? and how does one see that the substitution for $y(x)$ is 2-periodic?
$endgroup$
– Parseval
Jan 24 at 14:32












$begingroup$
$2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
$endgroup$
– Dylan
Jan 24 at 14:45




$begingroup$
$2$-periodic means it has a period of $2$, i.e. $y(x+2)=y(x)$. How did you obtain the Fourier series of $f(x)$ in your earlier question? How did you know it has terms in $e^{inpi x}$? How did you get through that question without knowing what it means?
$endgroup$
– Dylan
Jan 24 at 14:45












$begingroup$
Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
$endgroup$
– Parseval
Jan 24 at 14:56




$begingroup$
Okay, maybe my question was a bit misleading. I know what periodicity refers to, however in my book it says that one can express a function $f$ on the form $$f(x)=sum_{-infty}^{infty}c_n e^{i n x},$$ so I'm wondering why you substituted $$y(x)=c_0+sum_{-infty}^{infty}c_ne^{inpi x}.$$ Why do you latch on the $c_0$ when they excluded it in my book?
$endgroup$
– Parseval
Jan 24 at 14:56




1




1




$begingroup$
The first series does include a constant term. What is $e^{inx}$ when $n=0$?
$endgroup$
– Dylan
Jan 24 at 15:03




$begingroup$
The first series does include a constant term. What is $e^{inx}$ when $n=0$?
$endgroup$
– Dylan
Jan 24 at 15:03












$begingroup$
Ok I got it now! Thanks.
$endgroup$
– Parseval
Jan 24 at 15:08




$begingroup$
Ok I got it now! Thanks.
$endgroup$
– Parseval
Jan 24 at 15:08


















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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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