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Finding the fundamental solution of an ODE


Global solution for ODEHelp finding the particular solution of a 2nd order ODEODE time of life of a solutionNature of solution of ODE.General solution to ODEGeneral Solution of an ODESolution of first order nonlinear ODEHomogeneous second-order ODE with non-constant coefficientsLinear second order ODE, how model parameters change the solution?













0












$begingroup$


I want to find the fundamental solution of this ODE:
$$- u^primeprime + k^2 u=0, -infty<x<infty, kneq 0$$
I know that it is:
$$Gamma(x,epsilon) = frace^k2k$$, but I don’t know how to obtain it.



We know that for $xneq epsilon$
$$-Gamma^primeprime(x,epsilon) + k^2 Gamma(x,epsilon) =0$$, so we can say that for $xneq epsilon$
We have:
$$ Gamma(x,epsilon) = Ae^k(x-epsilon) + Be^-k(x-epsilon)$$
Can anyone please show me how can I compute A and B?



Thanks.










share|cite|improve this question









$endgroup$











  • $begingroup$
    $e^k$ is not a solution. By the way, why did you drop $u$ ?
    $endgroup$
    – Yves Daoust
    Mar 22 at 9:03











  • $begingroup$
    $$u(x)=c_1e^kx+c_2e^-kx$$
    $endgroup$
    – JJacquelin
    Mar 22 at 9:09










  • $begingroup$
    Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
    $endgroup$
    – LutzL
    Mar 22 at 11:36










  • $begingroup$
    You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
    $endgroup$
    – LutzL
    Mar 22 at 12:43










  • $begingroup$
    Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
    $endgroup$
    – LutzL
    Mar 22 at 14:26
















0












$begingroup$


I want to find the fundamental solution of this ODE:
$$- u^primeprime + k^2 u=0, -infty<x<infty, kneq 0$$
I know that it is:
$$Gamma(x,epsilon) = frace^k2k$$, but I don’t know how to obtain it.



We know that for $xneq epsilon$
$$-Gamma^primeprime(x,epsilon) + k^2 Gamma(x,epsilon) =0$$, so we can say that for $xneq epsilon$
We have:
$$ Gamma(x,epsilon) = Ae^k(x-epsilon) + Be^-k(x-epsilon)$$
Can anyone please show me how can I compute A and B?



Thanks.










share|cite|improve this question









$endgroup$











  • $begingroup$
    $e^k$ is not a solution. By the way, why did you drop $u$ ?
    $endgroup$
    – Yves Daoust
    Mar 22 at 9:03











  • $begingroup$
    $$u(x)=c_1e^kx+c_2e^-kx$$
    $endgroup$
    – JJacquelin
    Mar 22 at 9:09










  • $begingroup$
    Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
    $endgroup$
    – LutzL
    Mar 22 at 11:36










  • $begingroup$
    You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
    $endgroup$
    – LutzL
    Mar 22 at 12:43










  • $begingroup$
    Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
    $endgroup$
    – LutzL
    Mar 22 at 14:26














0












0








0





$begingroup$


I want to find the fundamental solution of this ODE:
$$- u^primeprime + k^2 u=0, -infty<x<infty, kneq 0$$
I know that it is:
$$Gamma(x,epsilon) = frace^k2k$$, but I don’t know how to obtain it.



We know that for $xneq epsilon$
$$-Gamma^primeprime(x,epsilon) + k^2 Gamma(x,epsilon) =0$$, so we can say that for $xneq epsilon$
We have:
$$ Gamma(x,epsilon) = Ae^k(x-epsilon) + Be^-k(x-epsilon)$$
Can anyone please show me how can I compute A and B?



Thanks.










share|cite|improve this question









$endgroup$




I want to find the fundamental solution of this ODE:
$$- u^primeprime + k^2 u=0, -infty<x<infty, kneq 0$$
I know that it is:
$$Gamma(x,epsilon) = frace^k2k$$, but I don’t know how to obtain it.



We know that for $xneq epsilon$
$$-Gamma^primeprime(x,epsilon) + k^2 Gamma(x,epsilon) =0$$, so we can say that for $xneq epsilon$
We have:
$$ Gamma(x,epsilon) = Ae^k(x-epsilon) + Be^-k(x-epsilon)$$
Can anyone please show me how can I compute A and B?



Thanks.







ordinary-differential-equations greens-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 8:56









Math12Math12

629




629











  • $begingroup$
    $e^k$ is not a solution. By the way, why did you drop $u$ ?
    $endgroup$
    – Yves Daoust
    Mar 22 at 9:03











  • $begingroup$
    $$u(x)=c_1e^kx+c_2e^-kx$$
    $endgroup$
    – JJacquelin
    Mar 22 at 9:09










  • $begingroup$
    Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
    $endgroup$
    – LutzL
    Mar 22 at 11:36










  • $begingroup$
    You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
    $endgroup$
    – LutzL
    Mar 22 at 12:43










  • $begingroup$
    Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
    $endgroup$
    – LutzL
    Mar 22 at 14:26

















  • $begingroup$
    $e^k$ is not a solution. By the way, why did you drop $u$ ?
    $endgroup$
    – Yves Daoust
    Mar 22 at 9:03











  • $begingroup$
    $$u(x)=c_1e^kx+c_2e^-kx$$
    $endgroup$
    – JJacquelin
    Mar 22 at 9:09










  • $begingroup$
    Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
    $endgroup$
    – LutzL
    Mar 22 at 11:36










  • $begingroup$
    You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
    $endgroup$
    – LutzL
    Mar 22 at 12:43










  • $begingroup$
    Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
    $endgroup$
    – LutzL
    Mar 22 at 14:26
















$begingroup$
$e^k$ is not a solution. By the way, why did you drop $u$ ?
$endgroup$
– Yves Daoust
Mar 22 at 9:03





$begingroup$
$e^k$ is not a solution. By the way, why did you drop $u$ ?
$endgroup$
– Yves Daoust
Mar 22 at 9:03













$begingroup$
$$u(x)=c_1e^kx+c_2e^-kx$$
$endgroup$
– JJacquelin
Mar 22 at 9:09




$begingroup$
$$u(x)=c_1e^kx+c_2e^-kx$$
$endgroup$
– JJacquelin
Mar 22 at 9:09












$begingroup$
Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
$endgroup$
– LutzL
Mar 22 at 11:36




$begingroup$
Your function is a solution to $-u''+k^2u=delta(x-ϵ)$ with the Dirac-delta distribution, otherwise connected to the concept of a Green's function. If you could provide more context on what you want to use this function for?
$endgroup$
– LutzL
Mar 22 at 11:36












$begingroup$
You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
$endgroup$
– LutzL
Mar 22 at 12:43




$begingroup$
You get a solution for $x<ϵ$, one for $x>ϵ$, both bounded, and have to connect them so that $u'(x)$ has a jump of height 1 at $x=ϵ$. This gives exactly your function.
$endgroup$
– LutzL
Mar 22 at 12:43












$begingroup$
Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
$endgroup$
– LutzL
Mar 22 at 14:26





$begingroup$
Could you cite your definition of "fundamental solution"? I can guess, but for an answer I'd like to be certain. Note that the fundamental or Wronskian matrix of the associated first order system might also sometimes be called fundamental solution.
$endgroup$
– LutzL
Mar 22 at 14:26











2 Answers
2






active

oldest

votes


















1












$begingroup$

This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $Gamma(x,epsilon)$ indeed is a linear combination of $e^k(x-epsilon)$ and $e^-k(x-epsilon)$ away from $x=epsilon$. However, one must remember that $Gamma$ is not differentiable for $x=epsilon$, so you should expect the general form of $Gamma$ to be
$$
Gamma(x,epsilon)=
begincases
A e^k(x-epsilon)+B e^-k(x-epsilon) & x<epsilon,,
\
C e^k(x-epsilon)+D e^-k(x-epsilon) & x>epsilon,.
endcases
$$

To find the coefficients $A,B,C,DinmathbbR$, you need to use the definition of fundamental solution. Namely, for any test function $varphiinmathcalC_c^infty(mathbbR)$ it must hold
beginequation
varphi(epsilon),=,int_-infty^inftyGamma(x-epsilon)left[-varphi''(x)+k^2varphi(x)right],dx,.
endequation

Split the integral on the right hand side into two parts (from $-infty$ to $epsilon-delta$ and from $epsilon+delta$ to $+infty$, with $delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $varphi(epsilon)$ at the limit $deltato 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $Gamma,'(x-epsilon)$ to have a jump of height $1$ at $x=epsilon$.
Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $Gamma$ is a fundamental solution, then $Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).



Good luck!






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    Write



    $$u'u''=k^2uu'$$ and integrate to get



    $$u'^2=k^2u^2pm c^2.$$



    This is a separable equation,



    $$fracu'sqrtdfrack^2c^2u^2pm1=c^2$$



    which integrates as



    $$frac cktextarcoshfrackuc=pm cx+c'$$ or



    $$frac cktextarsinhfrackuc=pm cx+c'.$$



    Finally,



    $$u=frac ckcosh(pm kx+c'')$$ or



    $$u=frac cksinh(pm kx+c'').$$



    You can convince yourself that this covers all cases of



    $$ae^kx+be^-kx.$$




    Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation



    $$lambda^2-k^2=0.$$




    As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $Gamma(x,epsilon)$ indeed is a linear combination of $e^k(x-epsilon)$ and $e^-k(x-epsilon)$ away from $x=epsilon$. However, one must remember that $Gamma$ is not differentiable for $x=epsilon$, so you should expect the general form of $Gamma$ to be
      $$
      Gamma(x,epsilon)=
      begincases
      A e^k(x-epsilon)+B e^-k(x-epsilon) & x<epsilon,,
      \
      C e^k(x-epsilon)+D e^-k(x-epsilon) & x>epsilon,.
      endcases
      $$

      To find the coefficients $A,B,C,DinmathbbR$, you need to use the definition of fundamental solution. Namely, for any test function $varphiinmathcalC_c^infty(mathbbR)$ it must hold
      beginequation
      varphi(epsilon),=,int_-infty^inftyGamma(x-epsilon)left[-varphi''(x)+k^2varphi(x)right],dx,.
      endequation

      Split the integral on the right hand side into two parts (from $-infty$ to $epsilon-delta$ and from $epsilon+delta$ to $+infty$, with $delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $varphi(epsilon)$ at the limit $deltato 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $Gamma,'(x-epsilon)$ to have a jump of height $1$ at $x=epsilon$.
      Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $Gamma$ is a fundamental solution, then $Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).



      Good luck!






      share|cite|improve this answer











      $endgroup$

















        1












        $begingroup$

        This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $Gamma(x,epsilon)$ indeed is a linear combination of $e^k(x-epsilon)$ and $e^-k(x-epsilon)$ away from $x=epsilon$. However, one must remember that $Gamma$ is not differentiable for $x=epsilon$, so you should expect the general form of $Gamma$ to be
        $$
        Gamma(x,epsilon)=
        begincases
        A e^k(x-epsilon)+B e^-k(x-epsilon) & x<epsilon,,
        \
        C e^k(x-epsilon)+D e^-k(x-epsilon) & x>epsilon,.
        endcases
        $$

        To find the coefficients $A,B,C,DinmathbbR$, you need to use the definition of fundamental solution. Namely, for any test function $varphiinmathcalC_c^infty(mathbbR)$ it must hold
        beginequation
        varphi(epsilon),=,int_-infty^inftyGamma(x-epsilon)left[-varphi''(x)+k^2varphi(x)right],dx,.
        endequation

        Split the integral on the right hand side into two parts (from $-infty$ to $epsilon-delta$ and from $epsilon+delta$ to $+infty$, with $delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $varphi(epsilon)$ at the limit $deltato 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $Gamma,'(x-epsilon)$ to have a jump of height $1$ at $x=epsilon$.
        Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $Gamma$ is a fundamental solution, then $Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).



        Good luck!






        share|cite|improve this answer











        $endgroup$















          1












          1








          1





          $begingroup$

          This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $Gamma(x,epsilon)$ indeed is a linear combination of $e^k(x-epsilon)$ and $e^-k(x-epsilon)$ away from $x=epsilon$. However, one must remember that $Gamma$ is not differentiable for $x=epsilon$, so you should expect the general form of $Gamma$ to be
          $$
          Gamma(x,epsilon)=
          begincases
          A e^k(x-epsilon)+B e^-k(x-epsilon) & x<epsilon,,
          \
          C e^k(x-epsilon)+D e^-k(x-epsilon) & x>epsilon,.
          endcases
          $$

          To find the coefficients $A,B,C,DinmathbbR$, you need to use the definition of fundamental solution. Namely, for any test function $varphiinmathcalC_c^infty(mathbbR)$ it must hold
          beginequation
          varphi(epsilon),=,int_-infty^inftyGamma(x-epsilon)left[-varphi''(x)+k^2varphi(x)right],dx,.
          endequation

          Split the integral on the right hand side into two parts (from $-infty$ to $epsilon-delta$ and from $epsilon+delta$ to $+infty$, with $delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $varphi(epsilon)$ at the limit $deltato 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $Gamma,'(x-epsilon)$ to have a jump of height $1$ at $x=epsilon$.
          Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $Gamma$ is a fundamental solution, then $Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).



          Good luck!






          share|cite|improve this answer











          $endgroup$



          This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $Gamma(x,epsilon)$ indeed is a linear combination of $e^k(x-epsilon)$ and $e^-k(x-epsilon)$ away from $x=epsilon$. However, one must remember that $Gamma$ is not differentiable for $x=epsilon$, so you should expect the general form of $Gamma$ to be
          $$
          Gamma(x,epsilon)=
          begincases
          A e^k(x-epsilon)+B e^-k(x-epsilon) & x<epsilon,,
          \
          C e^k(x-epsilon)+D e^-k(x-epsilon) & x>epsilon,.
          endcases
          $$

          To find the coefficients $A,B,C,DinmathbbR$, you need to use the definition of fundamental solution. Namely, for any test function $varphiinmathcalC_c^infty(mathbbR)$ it must hold
          beginequation
          varphi(epsilon),=,int_-infty^inftyGamma(x-epsilon)left[-varphi''(x)+k^2varphi(x)right],dx,.
          endequation

          Split the integral on the right hand side into two parts (from $-infty$ to $epsilon-delta$ and from $epsilon+delta$ to $+infty$, with $delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $varphi(epsilon)$ at the limit $deltato 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $Gamma,'(x-epsilon)$ to have a jump of height $1$ at $x=epsilon$.
          Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $Gamma$ is a fundamental solution, then $Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).



          Good luck!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 25 at 12:46

























          answered Mar 24 at 16:46









          Stefano BorghiniStefano Borghini

          114




          114





















              0












              $begingroup$

              Write



              $$u'u''=k^2uu'$$ and integrate to get



              $$u'^2=k^2u^2pm c^2.$$



              This is a separable equation,



              $$fracu'sqrtdfrack^2c^2u^2pm1=c^2$$



              which integrates as



              $$frac cktextarcoshfrackuc=pm cx+c'$$ or



              $$frac cktextarsinhfrackuc=pm cx+c'.$$



              Finally,



              $$u=frac ckcosh(pm kx+c'')$$ or



              $$u=frac cksinh(pm kx+c'').$$



              You can convince yourself that this covers all cases of



              $$ae^kx+be^-kx.$$




              Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation



              $$lambda^2-k^2=0.$$




              As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                Write



                $$u'u''=k^2uu'$$ and integrate to get



                $$u'^2=k^2u^2pm c^2.$$



                This is a separable equation,



                $$fracu'sqrtdfrack^2c^2u^2pm1=c^2$$



                which integrates as



                $$frac cktextarcoshfrackuc=pm cx+c'$$ or



                $$frac cktextarsinhfrackuc=pm cx+c'.$$



                Finally,



                $$u=frac ckcosh(pm kx+c'')$$ or



                $$u=frac cksinh(pm kx+c'').$$



                You can convince yourself that this covers all cases of



                $$ae^kx+be^-kx.$$




                Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation



                $$lambda^2-k^2=0.$$




                As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Write



                  $$u'u''=k^2uu'$$ and integrate to get



                  $$u'^2=k^2u^2pm c^2.$$



                  This is a separable equation,



                  $$fracu'sqrtdfrack^2c^2u^2pm1=c^2$$



                  which integrates as



                  $$frac cktextarcoshfrackuc=pm cx+c'$$ or



                  $$frac cktextarsinhfrackuc=pm cx+c'.$$



                  Finally,



                  $$u=frac ckcosh(pm kx+c'')$$ or



                  $$u=frac cksinh(pm kx+c'').$$



                  You can convince yourself that this covers all cases of



                  $$ae^kx+be^-kx.$$




                  Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation



                  $$lambda^2-k^2=0.$$




                  As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.






                  share|cite|improve this answer









                  $endgroup$



                  Write



                  $$u'u''=k^2uu'$$ and integrate to get



                  $$u'^2=k^2u^2pm c^2.$$



                  This is a separable equation,



                  $$fracu'sqrtdfrack^2c^2u^2pm1=c^2$$



                  which integrates as



                  $$frac cktextarcoshfrackuc=pm cx+c'$$ or



                  $$frac cktextarsinhfrackuc=pm cx+c'.$$



                  Finally,



                  $$u=frac ckcosh(pm kx+c'')$$ or



                  $$u=frac cksinh(pm kx+c'').$$



                  You can convince yourself that this covers all cases of



                  $$ae^kx+be^-kx.$$




                  Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation



                  $$lambda^2-k^2=0.$$




                  As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 22 at 9:25









                  Yves DaoustYves Daoust

                  132k676230




                  132k676230



























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