Ordinary differential equations “ode” [closed]












1












$begingroup$


Could you please help me to solve this inhomogeneous equation?



$y''(x)+xy'(x)+y(x)=g(x)$



where g only depends linearly on the approximation, in particular it will be a polynomial.



Let say $y(x)= y_{H}(x) +y_{p}(x)$
Where $y_{p}(x)$ is particular solution and $y_{H}(x)$ is linear combination of the fundamental solution. In the case that coefficients are constant we can use characteristic polynomial for finding fundamental solution but I don't know how can I find it in variable coefficients?!






ordinary-differential-equations







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$endgroup$



closed as off-topic by Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo Jan 24 at 0:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What denotes $$g$$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 18:41








  • 2




    $begingroup$
    @Dr.SonnhardGraubner it's a function
    $endgroup$
    – user637300
    Jan 22 at 18:46






  • 1




    $begingroup$
    @user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
    $endgroup$
    – Adrian Keister
    Jan 22 at 18:54






  • 1




    $begingroup$
    Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
    $endgroup$
    – Adrian Keister
    Jan 22 at 19:10






  • 1




    $begingroup$
    @AdrianKeister do you have any suggestions for solving this challenge?
    $endgroup$
    – user637300
    Jan 22 at 19:11
















1












$begingroup$


Could you please help me to solve this inhomogeneous equation?



$y''(x)+xy'(x)+y(x)=g(x)$



where g only depends linearly on the approximation, in particular it will be a polynomial.



Let say $y(x)= y_{H}(x) +y_{p}(x)$
Where $y_{p}(x)$ is particular solution and $y_{H}(x)$ is linear combination of the fundamental solution. In the case that coefficients are constant we can use characteristic polynomial for finding fundamental solution but I don't know how can I find it in variable coefficients?!






ordinary-differential-equations







share|cite|improve this question











$endgroup$



closed as off-topic by Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo Jan 24 at 0:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What denotes $$g$$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 18:41








  • 2




    $begingroup$
    @Dr.SonnhardGraubner it's a function
    $endgroup$
    – user637300
    Jan 22 at 18:46






  • 1




    $begingroup$
    @user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
    $endgroup$
    – Adrian Keister
    Jan 22 at 18:54






  • 1




    $begingroup$
    Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
    $endgroup$
    – Adrian Keister
    Jan 22 at 19:10






  • 1




    $begingroup$
    @AdrianKeister do you have any suggestions for solving this challenge?
    $endgroup$
    – user637300
    Jan 22 at 19:11














1












1








1


1



$begingroup$


Could you please help me to solve this inhomogeneous equation?



$y''(x)+xy'(x)+y(x)=g(x)$



where g only depends linearly on the approximation, in particular it will be a polynomial.



Let say $y(x)= y_{H}(x) +y_{p}(x)$
Where $y_{p}(x)$ is particular solution and $y_{H}(x)$ is linear combination of the fundamental solution. In the case that coefficients are constant we can use characteristic polynomial for finding fundamental solution but I don't know how can I find it in variable coefficients?!






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Could you please help me to solve this inhomogeneous equation?



$y''(x)+xy'(x)+y(x)=g(x)$



where g only depends linearly on the approximation, in particular it will be a polynomial.



Let say $y(x)= y_{H}(x) +y_{p}(x)$
Where $y_{p}(x)$ is particular solution and $y_{H}(x)$ is linear combination of the fundamental solution. In the case that coefficients are constant we can use characteristic polynomial for finding fundamental solution but I don't know how can I find it in variable coefficients?!






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:37







user637300

















asked Jan 22 at 18:38









user637300user637300

233




233




closed as off-topic by Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo Jan 24 at 0:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo Jan 24 at 0:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Pierce, Adrian Keister, Dylan, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    What denotes $$g$$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 18:41








  • 2




    $begingroup$
    @Dr.SonnhardGraubner it's a function
    $endgroup$
    – user637300
    Jan 22 at 18:46






  • 1




    $begingroup$
    @user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
    $endgroup$
    – Adrian Keister
    Jan 22 at 18:54






  • 1




    $begingroup$
    Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
    $endgroup$
    – Adrian Keister
    Jan 22 at 19:10






  • 1




    $begingroup$
    @AdrianKeister do you have any suggestions for solving this challenge?
    $endgroup$
    – user637300
    Jan 22 at 19:11














  • 1




    $begingroup$
    What denotes $$g$$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 18:41








  • 2




    $begingroup$
    @Dr.SonnhardGraubner it's a function
    $endgroup$
    – user637300
    Jan 22 at 18:46






  • 1




    $begingroup$
    @user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
    $endgroup$
    – Adrian Keister
    Jan 22 at 18:54






  • 1




    $begingroup$
    Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
    $endgroup$
    – Adrian Keister
    Jan 22 at 19:10






  • 1




    $begingroup$
    @AdrianKeister do you have any suggestions for solving this challenge?
    $endgroup$
    – user637300
    Jan 22 at 19:11








1




1




$begingroup$
What denotes $$g$$ here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 18:41






$begingroup$
What denotes $$g$$ here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 18:41






2




2




$begingroup$
@Dr.SonnhardGraubner it's a function
$endgroup$
– user637300
Jan 22 at 18:46




$begingroup$
@Dr.SonnhardGraubner it's a function
$endgroup$
– user637300
Jan 22 at 18:46




1




1




$begingroup$
@user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
$endgroup$
– Adrian Keister
Jan 22 at 18:54




$begingroup$
@user637300: I think Dr. Graubner meant which function is $g?$ Do you know what it is? If so, please edit the question to provide that information.
$endgroup$
– Adrian Keister
Jan 22 at 18:54




1




1




$begingroup$
Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
$endgroup$
– Adrian Keister
Jan 22 at 19:10




$begingroup$
Well, changing the problem from $y''+xy+1=g$ to $y''+xy'+y=g$ is a gigantic change!!
$endgroup$
– Adrian Keister
Jan 22 at 19:10




1




1




$begingroup$
@AdrianKeister do you have any suggestions for solving this challenge?
$endgroup$
– user637300
Jan 22 at 19:11




$begingroup$
@AdrianKeister do you have any suggestions for solving this challenge?
$endgroup$
– user637300
Jan 22 at 19:11










2 Answers
2






active

oldest

votes


















1












$begingroup$

The homogeneous equation is
$$y_h''+xy_h'+y_h=0,$$
with solution
$$y_h(x)=C_1e^{-x^2/2}+C_2e^{-x^2/2}int_0^xe^{t^2},dt,$$
obtained from the Wolfram Development Platform. As for a particular solution, if $g(x)$ is a polynomial of order $n,$ then we can see that the LHS, given an ansatz of a polynomial of order $n,$ will also be a polynomial of order $n.$ Therefore, I would try $y_p$ as a polynomial of order $n$ and work out the coefficients by plugging in, once $g$ is known.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Using the Laplace Transform, one obtains (assuming g is a polynomial with constant coefficients):



    $y(x) = e^{-frac{x^{2}}{2}} ,[a-i;b;sqrt{frac{pi}{2}};erf(frac{i;x}{sqrt{2}})+sum_{k=0}^nfrac{2^{frac{k}{2}};e^{-frac{pi,i}{2}(k+2)}}{k+1}g_{k},gamma(frac{k}{2}+1,-frac{x^{2}}{2})]$



    where $a = y(0)$, $b = y'(0)$, $g_{k}$ are the coefficients of the polynomial, $erf$ is the error function and $gamma$ is the incomplete gamma function






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      can you solve it for me?
      $endgroup$
      – user637300
      Jan 22 at 18:49






    • 1




      $begingroup$
      @user637300 the point of homework is to (at least try to) do it yourself
      $endgroup$
      – LoveTooNap29
      Jan 22 at 18:55






    • 1




      $begingroup$
      Yes I can give me a moment or 2
      $endgroup$
      – Fizzer
      Jan 22 at 18:55










    • $begingroup$
      Solved -- see above.
      $endgroup$
      – Fizzer
      Jan 23 at 2:26


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The homogeneous equation is
    $$y_h''+xy_h'+y_h=0,$$
    with solution
    $$y_h(x)=C_1e^{-x^2/2}+C_2e^{-x^2/2}int_0^xe^{t^2},dt,$$
    obtained from the Wolfram Development Platform. As for a particular solution, if $g(x)$ is a polynomial of order $n,$ then we can see that the LHS, given an ansatz of a polynomial of order $n,$ will also be a polynomial of order $n.$ Therefore, I would try $y_p$ as a polynomial of order $n$ and work out the coefficients by plugging in, once $g$ is known.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The homogeneous equation is
      $$y_h''+xy_h'+y_h=0,$$
      with solution
      $$y_h(x)=C_1e^{-x^2/2}+C_2e^{-x^2/2}int_0^xe^{t^2},dt,$$
      obtained from the Wolfram Development Platform. As for a particular solution, if $g(x)$ is a polynomial of order $n,$ then we can see that the LHS, given an ansatz of a polynomial of order $n,$ will also be a polynomial of order $n.$ Therefore, I would try $y_p$ as a polynomial of order $n$ and work out the coefficients by plugging in, once $g$ is known.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The homogeneous equation is
        $$y_h''+xy_h'+y_h=0,$$
        with solution
        $$y_h(x)=C_1e^{-x^2/2}+C_2e^{-x^2/2}int_0^xe^{t^2},dt,$$
        obtained from the Wolfram Development Platform. As for a particular solution, if $g(x)$ is a polynomial of order $n,$ then we can see that the LHS, given an ansatz of a polynomial of order $n,$ will also be a polynomial of order $n.$ Therefore, I would try $y_p$ as a polynomial of order $n$ and work out the coefficients by plugging in, once $g$ is known.






        share|cite|improve this answer









        $endgroup$



        The homogeneous equation is
        $$y_h''+xy_h'+y_h=0,$$
        with solution
        $$y_h(x)=C_1e^{-x^2/2}+C_2e^{-x^2/2}int_0^xe^{t^2},dt,$$
        obtained from the Wolfram Development Platform. As for a particular solution, if $g(x)$ is a polynomial of order $n,$ then we can see that the LHS, given an ansatz of a polynomial of order $n,$ will also be a polynomial of order $n.$ Therefore, I would try $y_p$ as a polynomial of order $n$ and work out the coefficients by plugging in, once $g$ is known.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 20:28









        Adrian KeisterAdrian Keister

        5,27371933




        5,27371933























            1












            $begingroup$

            Using the Laplace Transform, one obtains (assuming g is a polynomial with constant coefficients):



            $y(x) = e^{-frac{x^{2}}{2}} ,[a-i;b;sqrt{frac{pi}{2}};erf(frac{i;x}{sqrt{2}})+sum_{k=0}^nfrac{2^{frac{k}{2}};e^{-frac{pi,i}{2}(k+2)}}{k+1}g_{k},gamma(frac{k}{2}+1,-frac{x^{2}}{2})]$



            where $a = y(0)$, $b = y'(0)$, $g_{k}$ are the coefficients of the polynomial, $erf$ is the error function and $gamma$ is the incomplete gamma function






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              can you solve it for me?
              $endgroup$
              – user637300
              Jan 22 at 18:49






            • 1




              $begingroup$
              @user637300 the point of homework is to (at least try to) do it yourself
              $endgroup$
              – LoveTooNap29
              Jan 22 at 18:55






            • 1




              $begingroup$
              Yes I can give me a moment or 2
              $endgroup$
              – Fizzer
              Jan 22 at 18:55










            • $begingroup$
              Solved -- see above.
              $endgroup$
              – Fizzer
              Jan 23 at 2:26
















            1












            $begingroup$

            Using the Laplace Transform, one obtains (assuming g is a polynomial with constant coefficients):



            $y(x) = e^{-frac{x^{2}}{2}} ,[a-i;b;sqrt{frac{pi}{2}};erf(frac{i;x}{sqrt{2}})+sum_{k=0}^nfrac{2^{frac{k}{2}};e^{-frac{pi,i}{2}(k+2)}}{k+1}g_{k},gamma(frac{k}{2}+1,-frac{x^{2}}{2})]$



            where $a = y(0)$, $b = y'(0)$, $g_{k}$ are the coefficients of the polynomial, $erf$ is the error function and $gamma$ is the incomplete gamma function






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              can you solve it for me?
              $endgroup$
              – user637300
              Jan 22 at 18:49






            • 1




              $begingroup$
              @user637300 the point of homework is to (at least try to) do it yourself
              $endgroup$
              – LoveTooNap29
              Jan 22 at 18:55






            • 1




              $begingroup$
              Yes I can give me a moment or 2
              $endgroup$
              – Fizzer
              Jan 22 at 18:55










            • $begingroup$
              Solved -- see above.
              $endgroup$
              – Fizzer
              Jan 23 at 2:26














            1












            1








            1





            $begingroup$

            Using the Laplace Transform, one obtains (assuming g is a polynomial with constant coefficients):



            $y(x) = e^{-frac{x^{2}}{2}} ,[a-i;b;sqrt{frac{pi}{2}};erf(frac{i;x}{sqrt{2}})+sum_{k=0}^nfrac{2^{frac{k}{2}};e^{-frac{pi,i}{2}(k+2)}}{k+1}g_{k},gamma(frac{k}{2}+1,-frac{x^{2}}{2})]$



            where $a = y(0)$, $b = y'(0)$, $g_{k}$ are the coefficients of the polynomial, $erf$ is the error function and $gamma$ is the incomplete gamma function






            share|cite|improve this answer











            $endgroup$



            Using the Laplace Transform, one obtains (assuming g is a polynomial with constant coefficients):



            $y(x) = e^{-frac{x^{2}}{2}} ,[a-i;b;sqrt{frac{pi}{2}};erf(frac{i;x}{sqrt{2}})+sum_{k=0}^nfrac{2^{frac{k}{2}};e^{-frac{pi,i}{2}(k+2)}}{k+1}g_{k},gamma(frac{k}{2}+1,-frac{x^{2}}{2})]$



            where $a = y(0)$, $b = y'(0)$, $g_{k}$ are the coefficients of the polynomial, $erf$ is the error function and $gamma$ is the incomplete gamma function







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 23 at 2:13

























            answered Jan 22 at 18:46









            FizzerFizzer

            164




            164








            • 1




              $begingroup$
              can you solve it for me?
              $endgroup$
              – user637300
              Jan 22 at 18:49






            • 1




              $begingroup$
              @user637300 the point of homework is to (at least try to) do it yourself
              $endgroup$
              – LoveTooNap29
              Jan 22 at 18:55






            • 1




              $begingroup$
              Yes I can give me a moment or 2
              $endgroup$
              – Fizzer
              Jan 22 at 18:55










            • $begingroup$
              Solved -- see above.
              $endgroup$
              – Fizzer
              Jan 23 at 2:26














            • 1




              $begingroup$
              can you solve it for me?
              $endgroup$
              – user637300
              Jan 22 at 18:49






            • 1




              $begingroup$
              @user637300 the point of homework is to (at least try to) do it yourself
              $endgroup$
              – LoveTooNap29
              Jan 22 at 18:55






            • 1




              $begingroup$
              Yes I can give me a moment or 2
              $endgroup$
              – Fizzer
              Jan 22 at 18:55










            • $begingroup$
              Solved -- see above.
              $endgroup$
              – Fizzer
              Jan 23 at 2:26








            1




            1




            $begingroup$
            can you solve it for me?
            $endgroup$
            – user637300
            Jan 22 at 18:49




            $begingroup$
            can you solve it for me?
            $endgroup$
            – user637300
            Jan 22 at 18:49




            1




            1




            $begingroup$
            @user637300 the point of homework is to (at least try to) do it yourself
            $endgroup$
            – LoveTooNap29
            Jan 22 at 18:55




            $begingroup$
            @user637300 the point of homework is to (at least try to) do it yourself
            $endgroup$
            – LoveTooNap29
            Jan 22 at 18:55




            1




            1




            $begingroup$
            Yes I can give me a moment or 2
            $endgroup$
            – Fizzer
            Jan 22 at 18:55




            $begingroup$
            Yes I can give me a moment or 2
            $endgroup$
            – Fizzer
            Jan 22 at 18:55












            $begingroup$
            Solved -- see above.
            $endgroup$
            – Fizzer
            Jan 23 at 2:26




            $begingroup$
            Solved -- see above.
            $endgroup$
            – Fizzer
            Jan 23 at 2:26



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