{
  "index": "1951-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Two functions of \\( x \\) are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.",
  "solution": "Solution. We must find functions \\( f \\) and \\( g \\) such that\n\\[\n\\frac{f^{\\prime} g-f g^{\\prime}}{g^{2}}=\\frac{f^{\\prime}}{g^{\\prime}}\n\\]\nwhich is essentially equivalent to\n\\[\ng\\left(g^{\\prime}-g\\right) f^{\\prime}-g^{\\prime 2} f=0\n\\]\n\nNow, if we choose any interval \\( I \\) and any function \\( g \\) such that neither \\( g \\), nor \\( g^{\\prime}-g \\), nor \\( g^{\\prime} \\) vanishes on \\( I \\), then (2) is a non-singular first-order linear differential equation for \\( f \\) on \\( I \\) with the general solution\n\\[\nf(x)=C \\exp \\int^{x} \\frac{g^{\\prime 2}}{g\\left(g^{\\prime}-g\\right)} d t\n\\]\n\nWith any choice of \\( C \\neq 0 \\), the functions \\( f \\) and \\( g \\) will satisfy (1) and not vanish identically.\n\nTo be completely explicit, try \\( I=\\mathbf{R} \\) and \\( g(x)=e^{\\lambda x} \\). Then (2) becomes\n\\[\n(\\lambda-1) f^{\\prime}-\\lambda^{2} f=0\n\\]\nand we can take\n\\[\nf(x)=\\exp \\left(\\frac{\\lambda^{2}}{\\lambda-1}\\right) x\n\\]\nif \\( \\lambda \\neq 1 \\). A convenient choice is \\( \\lambda=2 \\). Then \\( f(x)=e^{4 x}, g(x)=e^{2 x} \\), \\( f(x) / g(x)=e^{2 x} \\) and\n\\[\n(f / g)^{\\prime}=2 e^{2 x}=f^{\\prime} / g^{\\prime}\n\\]\n\nRemark. Formula (3), with any choice of \\( g \\) such that \\( g\\left(g^{\\prime}-g\\right) g^{\\prime} \\) does not vanish on \\( I \\), gives the \"general solution\" of the problem. It does not, however, give all solutions because the requirement that \\( g^{\\prime}-g \\) not vanish on \\( I \\) is too restrictive. Thus\n\\[\n\\begin{array}{l}\nf(x)=x \\exp \\int_{0}^{x} \\frac{(t+2)^{2}}{1-t^{3}} d t \\\\\ng(x)=1+x+x^{2}\n\\end{array}\n\\]\nis a solution of the problem on \\( \\left(-\\frac{1}{2}, 1\\right) \\), but \\( g^{\\prime}-g \\) vanishes at 0 .",
  "vars": [
    "x",
    "t",
    "f",
    "g"
  ],
  "params": [
    "I",
    "C",
    "\\\\lambda"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "independentvariable",
        "t": "integrationvariable",
        "f": "firstfunction",
        "g": "secondfunction",
        "I": "intervalset",
        "C": "arbitraryconstant",
        "\\lambda": "lambdaparam"
      },
      "question": "2. Two functions of \\( independentvariable \\) are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.",
      "solution": "Solution. We must find functions \\( firstfunction \\) and \\( secondfunction \\) such that\n\\[\n\\frac{firstfunction^{\\prime} secondfunction-firstfunction secondfunction^{\\prime}}{secondfunction^{2}}=\\frac{firstfunction^{\\prime}}{secondfunction^{\\prime}}\n\\]\nwhich is essentially equivalent to\n\\[\nsecondfunction\\left(secondfunction^{\\prime}-secondfunction\\right) firstfunction^{\\prime}-secondfunction^{\\prime 2} firstfunction=0\n\\]\n\nNow, if we choose any interval \\( intervalset \\) and any function \\( secondfunction \\) such that neither \\( secondfunction \\), nor \\( secondfunction^{\\prime}-secondfunction \\), nor \\( secondfunction^{\\prime} \\) vanishes on \\( intervalset \\), then (2) is a non-singular first-order linear differential equation for \\( firstfunction \\) on \\( intervalset \\) with the general solution\n\\[\nfirstfunction(independentvariable)=arbitraryconstant \\exp \\int^{independentvariable} \\frac{secondfunction^{\\prime 2}}{secondfunction\\left(secondfunction^{\\prime}-secondfunction\\right)} d integrationvariable\n\\]\n\nWith any choice of \\( arbitraryconstant \\neq 0 \\), the functions \\( firstfunction \\) and \\( secondfunction \\) will satisfy (1) and not vanish identically.\n\nTo be completely explicit, try \\( intervalset=\\mathbf{R} \\) and \\( secondfunction(independentvariable)=e^{lambdaparam independentvariable} \\). Then (2) becomes\n\\[\n(lambdaparam-1) firstfunction^{\\prime}-lambdaparam^{2} firstfunction=0\n\\]\nand we can take\n\\[\nfirstfunction(independentvariable)=\\exp \\left(\\frac{lambdaparam^{2}}{lambdaparam-1}\\right) independentvariable\n\\]\nif \\( lambdaparam \\neq 1 \\). A convenient choice is \\( lambdaparam=2 \\). Then \\( firstfunction(independentvariable)=e^{4 independentvariable},\\; secondfunction(independentvariable)=e^{2 independentvariable} \\), \\( firstfunction(independentvariable) / secondfunction(independentvariable)=e^{2 independentvariable} \\) and\n\\[\n(firstfunction / secondfunction)^{\\prime}=2 e^{2 independentvariable}=firstfunction^{\\prime} / secondfunction^{\\prime}\n\\]\n\nRemark. Formula (3), with any choice of \\( secondfunction \\) such that \\( secondfunction\\left(secondfunction^{\\prime}-secondfunction\\right) secondfunction^{\\prime} \\) does not vanish on \\( intervalset \\), gives the \"general solution\" of the problem. It does not, however, give all solutions because the requirement that \\( secondfunction^{\\prime}-secondfunction \\) not vanish on \\( intervalset \\) is too restrictive. Thus\n\\[\n\\begin{array}{l}\nfirstfunction(independentvariable)=independentvariable \\exp \\int_{0}^{independentvariable} \\frac{(integrationvariable+2)^{2}}{1-integrationvariable^{3}} d integrationvariable \\\\\nsecondfunction(independentvariable)=1+independentvariable+independentvariable^{2}\n\\end{array}\n\\]\nis a solution of the problem on \\( \\left(-\\frac{1}{2}, 1\\right) \\), but \\( secondfunction^{\\prime}-secondfunction \\) vanishes at 0 ."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "longitude",
        "t": "sandwich",
        "f": "envelope",
        "g": "marigold",
        "I": "staircase",
        "C": "blueberry",
        "\\lambda": "astronaut"
      },
      "question": "2. Two functions of \\( longitude \\) are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.",
      "solution": "Solution. We must find functions \\( envelope \\) and \\( marigold \\) such that\n\\[\n\\frac{envelope^{\\prime} marigold-envelope marigold^{\\prime}}{marigold^{2}}=\\frac{envelope^{\\prime}}{marigold^{\\prime}}\n\\]\nwhich is essentially equivalent to\n\\[\nmarigold\\left(marigold^{\\prime}-marigold\\right) envelope^{\\prime}-marigold^{\\prime 2} envelope=0\n\\]\n\nNow, if we choose any interval \\( staircase \\) and any function \\( marigold \\) such that neither \\( marigold \\), nor \\( marigold^{\\prime}-marigold \\), nor \\( marigold^{\\prime} \\) vanishes on \\( staircase \\), then (2) is a non-singular first-order linear differential equation for \\( envelope \\) on \\( staircase \\) with the general solution\n\\[\nenvelope(longitude)=blueberry \\exp \\int^{longitude} \\frac{marigold^{\\prime 2}}{marigold\\left(marigold^{\\prime}-marigold\\right)} d sandwich\n\\]\n\nWith any choice of \\( blueberry \\neq 0 \\), the functions \\( envelope \\) and \\( marigold \\) will satisfy (1) and not vanish identically.\n\nTo be completely explicit, try \\( staircase=\\mathbf{R} \\) and \\( marigold(longitude)=e^{astronaut longitude} \\). Then (2) becomes\n\\[\n(astronaut-1) envelope^{\\prime}-astronaut^{2} envelope=0\n\\]\nand we can take\n\\[\nenvelope(longitude)=\\exp \\left(\\frac{astronaut^{2}}{astronaut-1}\\right) longitude\n\\]\nif \\( astronaut \\neq 1 \\). A convenient choice is \\( astronaut=2 \\). Then \\( envelope(longitude)=e^{4 longitude}, marigold(longitude)=e^{2 longitude} \\), \\( envelope(longitude) / marigold(longitude)=e^{2 longitude} \\) and\n\\[\n(envelope / marigold)^{\\prime}=2 e^{2 longitude}=envelope^{\\prime} / marigold^{\\prime}\n\\]\n\nRemark. Formula (3), with any choice of \\( marigold \\) such that \\( marigold\\left(marigold^{\\prime}-marigold\\right) marigold^{\\prime} \\) does not vanish on \\( staircase \\), gives the \"general solution\" of the problem. It does not, however, give all solutions because the requirement that \\( marigold^{\\prime}-marigold \\) not vanish on \\( staircase \\) is too restrictive. Thus\n\\[\n\\begin{array}{l}\nenvelope(longitude)=longitude \\exp \\int_{0}^{longitude} \\frac{(sandwich+2)^{2}}{1-sandwich^{3}} d sandwich \\\\\nmarigold(longitude)=1+longitude+longitude^{2}\n\\end{array}\n\\]\nis a solution of the problem on \\( \\left(-\\frac{1}{2}, 1\\right) \\), but \\( marigold^{\\prime}-marigold \\) vanishes at 0 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "outputvalue",
        "t": "staticquantity",
        "f": "nonfunction",
        "g": "constantval",
        "I": "discretept",
        "C": "variableval",
        "\\lambda": "mutableparm"
      },
      "question": "2. Two functions of \\( outputvalue \\) are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.",
      "solution": "Solution. We must find functions \\( nonfunction \\) and \\( constantval \\) such that\n\\[\n\\frac{nonfunction^{\\prime} constantval-nonfunction constantval^{\\prime}}{constantval^{2}}=\\frac{nonfunction^{\\prime}}{constantval^{\\prime}}\n\\]\nwhich is essentially equivalent to\n\\[\nconstantval\\left(constantval^{\\prime}-constantval\\right) nonfunction^{\\prime}-constantval^{\\prime 2} nonfunction=0\n\\]\n\nNow, if we choose any interval \\( discretept \\) and any function \\( constantval \\) such that neither \\( constantval \\), nor \\( constantval^{\\prime}-constantval \\), nor \\( constantval^{\\prime} \\) vanishes on \\( discretept \\), then (2) is a non-singular first-order linear differential equation for \\( nonfunction \\) on \\( discretept \\) with the general solution\n\\[\nnonfunction(outputvalue)=variableval \\exp \\int^{outputvalue} \\frac{constantval^{\\prime 2}}{constantval\\left(constantval^{\\prime}-constantval\\right)} d staticquantity\n\\]\n\nWith any choice of \\( variableval \\neq 0 \\), the functions \\( nonfunction \\) and \\( constantval \\) will satisfy (1) and not vanish identically.\n\nTo be completely explicit, try \\( discretept=\\mathbf{R} \\) and \\( constantval(outputvalue)=e^{mutableparm outputvalue} \\). Then (2) becomes\n\\[\n(mutableparm-1) nonfunction^{\\prime}-mutableparm^{2} nonfunction=0\n\\]\nand we can take\n\\[\nnonfunction(outputvalue)=\\exp \\left(\\frac{mutableparm^{2}}{mutableparm-1}\\right) outputvalue\n\\]\nif \\( mutableparm \\neq 1 \\). A convenient choice is \\( mutableparm=2 \\). Then \\( nonfunction(outputvalue)=e^{4 outputvalue}, constantval(outputvalue)=e^{2 outputvalue} \\), \\( nonfunction(outputvalue) / constantval(outputvalue)=e^{2 outputvalue} \\) and\n\\[\n(nonfunction / constantval)^{\\prime}=2 e^{2 outputvalue}=nonfunction^{\\prime} / constantval^{\\prime}\n\\]\n\nRemark. Formula (3), with any choice of \\( constantval \\) such that \\( constantval\\left(constantval^{\\prime}-constantval\\right) constantval^{\\prime} \\) does not vanish on \\( discretept \\), gives the \"general solution\" of the problem. It does not, however, give all solutions because the requirement that \\( constantval^{\\prime}-constantval \\) not vanish on \\( discretept \\) is too restrictive. Thus\n\\[\n\\begin{array}{l}\nnonfunction(outputvalue)=outputvalue \\exp \\int_{0}^{outputvalue} \\frac{(staticquantity+2)^{2}}{1-staticquantity^{3}} d staticquantity \\\\\nconstantval(outputvalue)=1+outputvalue+outputvalue^{2}\n\\end{array}\n\\]\nis a solution of the problem on \\( \\left(-\\frac{1}{2}, 1\\right) \\), but \\( constantval^{\\prime}-constantval \\) vanishes at 0 ."
    },
    "garbled_string": {
      "map": {
        "x": "qbvtuklef",
        "t": "jghsionva",
        "f": "ozpqwnmka",
        "g": "hzrctupori",
        "I": "mgwlaspde",
        "C": "xfczmnovq",
        "\\lambda": "qdpsnmytw"
      },
      "question": "2. Two functions of \\( qbvtuklef \\) are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.",
      "solution": "Solution. We must find functions \\( ozpqwnmka \\) and \\( hzrctupori \\) such that\n\\[\n\\frac{ozpqwnmka^{\\prime} hzrctupori-ozpqwnmka hzrctupori^{\\prime}}{hzrctupori^{2}}=\\frac{ozpqwnmka^{\\prime}}{hzrctupori^{\\prime}}\n\\]\nwhich is essentially equivalent to\n\\[\nhzrctupori\\left(hzrctupori^{\\prime}-hzrctupori\\right) ozpqwnmka^{\\prime}-hzrctupori^{\\prime 2} ozpqwnmka=0\n\\]\n\nNow, if we choose any interval \\( mgwlaspde \\) and any function \\( hzrctupori \\) such that neither \\( hzrctupori \\), nor \\( hzrctupori^{\\prime}-hzrctupori \\), nor \\( hzrctupori^{\\prime} \\) vanishes on \\( mgwlaspde \\), then (2) is a non-singular first-order linear differential equation for \\( ozpqwnmka \\) on \\( mgwlaspde \\) with the general solution\n\\[\nozpqwnmka(qbvtuklef)=xfczmnovq \\exp \\int^{qbvtuklef} \\frac{hzrctupori^{\\prime 2}}{hzrctupori\\left(hzrctupori^{\\prime}-hzrctupori\\right)} d jghsionva\n\\]\n\nWith any choice of \\( xfczmnovq \\neq 0 \\), the functions \\( ozpqwnmka \\) and \\( hzrctupori \\) will satisfy (1) and not vanish identically.\n\nTo be completely explicit, try \\( mgwlaspde=\\mathbf{R} \\) and \\( hzrctupori(qbvtuklef)=e^{qdpsnmytw qbvtuklef} \\). Then (2) becomes\n\\[\n(qdpsnmytw-1) ozpqwnmka^{\\prime}-qdpsnmytw^{2} ozpqwnmka=0\n\\]\nand we can take\n\\[\nozpqwnmka(qbvtuklef)=\\exp \\left(\\frac{qdpsnmytw^{2}}{qdpsnmytw-1}\\right) qbvtuklef\n\\]\nif \\( qdpsnmytw \\neq 1 \\). A convenient choice is \\( qdpsnmytw=2 \\). Then \\( ozpqwnmka(qbvtuklef)=e^{4 qbvtuklef}, hzrctupori(qbvtuklef)=e^{2 qbvtuklef} \\), \\( ozpqwnmka(qbvtuklef) / hzrctupori(qbvtuklef)=e^{2 qbvtuklef} \\) and\n\\[\n(ozpqwnmka / hzrctupori)^{\\prime}=2 e^{2 qbvtuklef}=ozpqwnmka^{\\prime} / hzrctupori^{\\prime}\n\\]\n\nRemark. Formula (3), with any choice of \\( hzrctupori \\) such that \\( hzrctupori\\left(hzrctupori^{\\prime}-hzrctupori\\right) hzrctupori^{\\prime} \\) does not vanish on \\( mgwlaspde \\), gives the \"general solution\" of the problem. It does not, however, give all solutions because the requirement that \\( hzrctupori^{\\prime}-hzrctupori \\) not vanish on \\( mgwlaspde \\) is too restrictive. Thus\n\\[\n\\begin{array}{l}\nozpqwnmka(qbvtuklef)=qbvtuklef \\exp \\int_{0}^{qbvtuklef} \\frac{(jghsionva+2)^{2}}{1-jghsionva^{3}} d jghsionva \\\\\nhzrctupori(qbvtuklef)=1+qbvtuklef+qbvtuklef^{2}\n\\end{array}\n\\]\nis a solution of the problem on \\( \\left(-\\frac{1}{2}, 1\\right) \\), but \\( hzrctupori^{\\prime}-hzrctupori \\) vanishes at 0 ."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nU=(0,\\infty)\\times(0,\\infty)\\subset \\mathbb R^{2},\\qquad  \nf,g\\in C^{\\omega}(U,\\mathbb R)\\;(=\\text{ real–analytic on }U).\n\\]\n\nDetermine \\emph{all} ordered pairs $(f,g)$ that satisfy  \n\n(i)  the five analytic functions  \n\\[\nf,\\;g,\\;\\partial_{x}g,\\;\\partial_{y}g,\\;\\partial^{2}_{xy}g\n\\]\nnever vanish on $U$;\n\n(ii)  for every $(x,y)\\in U$\n\\[\n\\partial_{x}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial_{x}f}{\\partial_{x}g},\\qquad\n\\partial_{y}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial_{y}f}{\\partial_{y}g},\\qquad\n\\partial^{2}_{xy}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial^{2}_{xy}f}{\\partial^{2}_{xy}g};\n\\]\n\n(iii) $f$ is \\emph{not} a constant multiple of $g$.\n\nEvery admissible couple $(f,g)$ belongs to one and only one of the three\nmutually disjoint classes below.\n\n(A) Separable family.  \nThere exist a base point $(x_{0},y_{0})\\in U$, a constant $C\\neq0$ and\nreal–analytic functions  \n\\[\nr:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0,1\\},\\qquad\ns:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0,1\\},\n\\]\nsuch that  \n\\[\n(s-1)^{2}\\not\\equiv(r-1)^{2}\\qquad\n\\bigl(\\text{so that families (A) and (B) remain disjoint}\\bigr),\n\\]\nand  \n\\[\ng(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x} r(t)\\,dt+\n               \\int_{y_{0}}^{y}s(u)\\,du\\Bigr),\\qquad\nf(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x}\\frac{r(t)^{2}}{r(t)-1}\\,dt+\n               \\int_{y_{0}}^{y}\\frac{s(u)^{2}}{s(u)-1}\\,du\\Bigr).\n\\]\n\n(B) Diagonal family.  \nThere exist $C\\neq0$ and a real–analytic function  \n\\[\nh:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0\\},\n\\]\nsuch that for every $t>0$\n\\[\nh'(t)\\neq0,\\qquad h''(t)\\neq0,\\qquad h'(t)-h(t)\\neq0,\n\\]\nand  \n\\[\ng(x,y)=C\\,h(x+y),\\qquad\nf(x,y)=C\\,h(x+y)\\,\n        \\exp\\!\\Bigl(\\int_{\\,x_{0}+y_{0}}^{\\,x+y}\n                    \\frac{h'(t)}{h'(t)-h(t)}\\,dt\\Bigr),\n\\]\nwhere $(x_{0},y_{0})\\in U$ is fixed once and for all.\n\n(C) Skew–exponential family.  \nThere exist $C\\neq0$, $\\tau_{0}\\in\\mathbb R$ and a real–analytic function  \n\\[\n\\varphi:\\mathbb R\\longrightarrow\\mathbb R\\setminus\\{0\\},\n\\]\nsuch that for every $t\\in\\mathbb R$\n\\[\n\\varphi(t)\\neq0,\\quad\\varphi'(t)\\neq0,\\quad\n\\varphi(t)+\\varphi'(t)\\neq0,\\quad\n2\\varphi(t)+\\varphi'(t)\\neq0,\\quad\n2\\varphi'(t)+\\varphi''(t)\\neq0,\n\\]\nand  \n\\[\ng(x,y)=C\\,\\mathrm e^{2x}\\,\\varphi(x-y),\\qquad\nf(x,y)=C\\,\\mathrm e^{2x}\\,\\varphi(x-y)\\,\n        \\exp\\!\\Bigl(2y+\\int_{\\tau_{0}}^{\\,x-y}\n                     \\frac{2\\varphi(t)+\\varphi'(t)}\n                          {\\varphi(t)+\\varphi'(t)}\\,dt\\Bigr).\n\\]\n\nConversely, every choice of data allowed in \\text{\\rm(A)}, \\text{\\rm(B)} \nor \\text{\\rm(C)} produces an admissible pair, and no other pairs exist.\n\nFinally, exhibit an explicit admissible pair that belongs to\n\\text{\\rm(C)} but to neither \\text{\\rm(A)} nor \\text{\\rm(B)}.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we set  \n\\[\nu:=\\frac{f}{g},\\qquad\ng_{x}:=\\partial_{x}g,\\;g_{y}:=\\partial_{y}g,\\;\ng_{xy}:=\\partial^{2}_{xy}g,\n\\]\nand use analogous notations for $f$ and $u$.  All derivatives are\ncomputed on $U$.\n\nBecause $f$ and $g$ are real–analytic, every expression constructed from\nthem is real–analytic.  This will be essential in Step~2.\n\n\\medskip\n\\noindent\\textbf{Remark on divisions.}\nConditions (i) already tell us $g_{x},g_{y}\\neq0$.  \nIn the course of the proof we shall divide by $g_{x}-g$ and\n$g_{y}-g$.  It will soon be shown (see \\eqref{eq:dichotomy}) that at\nevery point of $U$ at least one of $g_{x}-g$ and $g_{y}-g$ is\nnon–vanishing, and later that each admissible pair actually satisfies\nboth $g_{x}-g\\neq0$ and $g_{y}-g\\neq0$ everywhere.  Hence all divisions\nare legitimate.\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{1. First–order identities coming from \\textup{(ii)}.}\n%---------------------------------------------------------------------\nRelations (ii) give  \n\\[\nu_{x}=\\frac{f_{x}}{g_{x}},\\qquad\nu_{y}=\\frac{f_{y}}{g_{y}}.\\tag{1}\n\\]\nSince $f=ug$, differentiating and combining with (1) yields  \n\\[\nu\\,g_{x}=u_{x}\\,(g_{x}-g),\\qquad\nu\\,g_{y}=u_{y}\\,(g_{y}-g).\\tag{2}\n\\]\nAs $g_{x},g_{y}\\neq0$ we introduce the real–analytic ratios  \n\\[\nP:=\\frac{u_{x}}{u}=\\frac{g_{x}}{g_{x}-g},\\qquad\nQ:=\\frac{u_{y}}{u}=\\frac{g_{y}}{g_{y}-g}.\\tag{3}\n\\]\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{2. Two fundamental compatibility relations.}\n%---------------------------------------------------------------------\n(a)  Mixed partials of $u$.  \nTaking $\\partial_{y}$ of $P$ and $\\partial_{x}$ of $Q$, using\n$u_{xy}=u_{yx}$ and \\eqref{eq:dichotomy} one obtains  \n\\[\nP_{y}-Q_{x}= \n\\frac{-g\\,g_{xy}+g_{x}g_{y}}\n     {(g_{x}-g)^{2}(g_{y}-g)^{2}}\n\\Bigl[(g_{y}-g)^{2}-(g_{x}-g)^{2}\\Bigr].\n\\]\nSet  \n\\[\nA:=-g\\,g_{xy}+g_{x}g_{y},\\qquad\nB:=(g_{y}-g)^{2}-(g_{x}-g)^{2}.\\tag{4}\n\\]\nThus $u_{xy}=u_{yx}$ is equivalent to  \n\\[\nA\\,B\\equiv0.\\tag{5}\n\\]\n\n(b)  The $xy$–condition in (ii).  \nDifferentiating $f=ug$ twice and substituting into the\nthird equality in (ii) again leads, after simplification, to the same\nidentity $A\\,B\\equiv0$.  Hence (5) holds on $U$.\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{3. Analytic factorisation yields a dichotomy.}\n%---------------------------------------------------------------------\nBoth $A$ and $B$ are real–analytic, hence so is their product.  Since\n$U$ is connected and $A\\,B\\equiv0$, the identity principle gives  \n\\[\n\\boxed{A\\equiv0\\quad\\text{on }U\\qquad\\text{or}\\qquad B\\equiv0\\quad\\text{on }U.}\\tag{6}\n\\]\nThis global dichotomy will be the backbone of the classification.\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{4. Case $A\\equiv0$ – separable solutions (family \\textup{(A)}).}\n%---------------------------------------------------------------------\nIntroduce  \n\\[\nr:=\\frac{g_{x}}{g},\\qquad s:=\\frac{g_{y}}{g}.\n\\]\nBecause $A=-g\\,g_{xy}+g_{x}g_{y}\\equiv0$ we have $r_{y}=0$ and\n$s_{x}=0$, so $r=r(x)$ and $s=s(y)$.  Solving the system  \n\\[\n\\partial_{x}\\log g=r(x),\\qquad \\partial_{y}\\log g=s(y)\n\\]\nalong any path from $(x_{0},y_{0})$ to $(x,y)$ gives  \n\\[\ng(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x} r(t)\\,dt+\n               \\int_{y_{0}}^{y}s(u)\\,du\\Bigr).\\tag{7}\n\\]\n\nFormulae (2)–(3) yield  \n\\[\n\\frac{u_{x}}{u}=\\frac{r}{r-1},\\qquad\n\\frac{u_{y}}{u}=\\frac{s}{s-1},\n\\]\nwhence  \n\\[\nf(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x}\\frac{r(t)^{2}}{r(t)-1}\\,dt+\n               \\int_{y_{0}}^{y}\\frac{s(u)^{2}}{s(u)-1}\\,du\\Bigr).\\tag{8}\n\\]\n\nBecause $g_{xy}=r(x)s(y)g(x,y)$, the non–vanishing of $g_{xy}$ forces  \n\\[\nr(x)\\neq0\\ (\\forall x>0),\\qquad s(y)\\neq0\\ (\\forall y>0).\\tag{9}\n\\]\nWith (7)–(9) we recover exactly family (A).  The extra requirement\n$(s-1)^{2}\\not\\equiv(r-1)^{2}$ was inserted in the statement (not for\ncondition~(iii)) but to ensure that no separable solution coincides with\na forthcoming diagonal one.\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{5. Case $A\\not\\equiv0$ – hence $B\\equiv0$.}\n%---------------------------------------------------------------------\nBecause $g_{x},g_{y}\\neq0$, $B\\equiv0$ rewrites  \n\\[\n(g_{y}-g)^{2}\\equiv(g_{x}-g)^{2}.\n\\]\nHence the ratio $(g_{y}-g)/(g_{x}-g)$ is the \\emph{constant} $\\pm1$.\nTwo sub–cases arise.\n\n\\medskip\n\\noindent\\textbf{5.1 Sub–case $g_{y}=g_{x}$ – diagonal solutions (family \\textup{(B)}).}\n\nSet $s:=x+y$, $t:=x-y$.  The equality $g_{y}=g_{x}$ is $\\partial_{t}g=0$,\nso $g(x,y)=h(s)$ for some analytic $h\\neq0$.  Relations (2)–(3) imply\nthat $u$ also depends on $s$ only.  Integrating in $s$ gives  \n\\[\nf(x,y)=h(s)\\exp\\!\\Bigl(\\int_{s_{0}}^{s}\n                        \\frac{h'(v)}{h'(v)-h(v)}\\,dv\\Bigr),\\tag{10}\n\\]\nand the inequalities required in (i) translate into\n\\[\nh'\\neq0,\\quad h''\\neq0,\\quad h'-h\\neq0,\n\\]\nyielding family (B).\n\n\\medskip\n\\noindent\\textbf{5.2 Sub–case $g_{y}+g_{x}=2g$ – skew–exponential solutions (family \\textup{(C)}).}\n\nSolve the linear PDE $g_{x}+g_{y}=2g$ by characteristics.  Writing\n$s:=x-y$ one finds  \n\\[\ng(x,y)=\\Phi(s)\\,\\mathrm e^{2x},\\qquad \\Phi\\in C^{\\omega}(\\mathbb R),\\;\n\\Phi\\neq0.\\tag{11}\n\\]\nFormulae (2)–(3) give  \n\\[\n\\frac{u_{x}}{u}=p(s):=\\frac{2\\Phi+\\Phi'}{\\Phi+\\Phi'},\\qquad\n\\frac{u_{y}}{u}=q(s):=\\frac{\\Phi'}{\\Phi+\\Phi'},\n\\]\nso $p+q\\equiv2$ and $p'+q'=0$.  The system\n$u_{x}=p(s)u,\\;u_{y}=q(s)u$ is compatible and integrates to  \n\\[\nu(x,y)=D\\,\\mathrm e^{2y}\\,\n        \\exp\\!\\Bigl(\\int_{\\tau_{0}}^{s} p(t)\\,dt\\Bigr).\\tag{12}\n\\]\nWriting $\\varphi:=\\Phi$ and $C:=D$ produces exactly family (C).  The\nfive inequalities imposed on $\\varphi$ guarantee the non–vanishing\nconditions in (i).\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{6. Exhaustiveness and pair–wise disjointness.}\n%---------------------------------------------------------------------\nStep~3 provides the global dichotomy \\eqref{eq:dichotomy}.  The branch\n$A\\equiv0$ gives family (A); the branch $B\\equiv0$ splits into the\nexclusive alternatives $g_{x}=g_{y}$ and $g_{x}+g_{y}=2g$, yielding\nfamilies (B) and (C) respectively.  Conversely, direct substitution\nshows that every triple of data admitted in (A), (B) or (C) satisfies\nconditions (i)–(iii).  The additional requirement\n$(s-1)^{2}\\not\\equiv(r-1)^{2}$ prevents overlap between (A) and (B),\nwhile the sign conditions in (B) and (C) exclude overlap with the\nremaining family.  Hence the trichotomy is complete and the\nclassification unique.\n\n\\bigskip\n%---------------------------------------------------------------------\n\\textbf{7. An explicit pair belonging to \\textup{(C)} only.}\n%---------------------------------------------------------------------\nTake  \n\\[\n\\varphi(t):=1+\\mathrm e^{t},\\qquad\nC:=1,\\qquad\n\\tau_{0}:=0.\n\\]\nThen  \n\\[\n\\varphi>0,\\;\n\\varphi'>0,\\;\n\\varphi+\\varphi' = 1+2\\mathrm e^{t}>0,\\;\n2\\varphi+\\varphi' = 2+3\\mathrm e^{t}>0,\\;\n2\\varphi'+\\varphi'' = 3\\mathrm e^{t}>0,\n\\]\nso all imposed inequalities hold.  The resulting pair is  \n\\[\ng(x,y)=\\mathrm e^{2x}\\bigl(1+\\mathrm e^{x-y}\\bigr),\\qquad\nf(x,y)=\\mathrm e^{4x}\\,\n        \\frac{1+\\mathrm e^{x-y}}{\\sqrt{1+2\\mathrm e^{x-y}}}.\n\\]\nA direct computation confirms (ii) and the five non–vanishing\nconditions; moreover the solution is obviously neither separable nor\ndiagonal, so it belongs exclusively to family (C).\\qed\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.440516",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension: the problem moves from one variable to two variables, forcing the use of partial derivatives and mixed partials.\n2.  Additional constraints: three simultaneous identities (two first-order and one second-order) must hold, greatly restricting admissible functions.\n3.  Sophisticated structure: solving requires analysing an over-determined nonlinear PDE system, reducing it to an algebraic system, and checking global non-vanishing conditions.\n4.  Deeper theory: the exhaustiveness proof invokes separation of variables, the structure of solutions of the original ODE, and compatibility of cross-derivatives.\n5.  Multiple interacting concepts: one must synchronise conditions in x and y directions and for mixed derivatives; any naive “copy” of the 1-D example fails unless the parameters are tuned to satisfy all three algebraic equations.\n\nHence the variant is substantially harder than both the original problem and the current kernel variant, demanding multi-variable calculus, PDE reasoning, and a careful algebraic compatibility analysis."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nU=(0,\\infty)\\times(0,\\infty)\\subset\\mathbb R^{2},\n\\qquad  \nf,g\\in C^{\\omega}(U,\\mathbb R)\\;(=\\hbox{ real–analytic on }U).\n\\]\n\nDetermine \\emph{all} pairs $(f,g)$ that satisfy  \n\n(i)\\; $f,\\;g,\\;\\partial_{x}g,\\;\\partial_{y}g,\\;\\partial^{2}_{xy}g$ never vanish on $U$;\n\n(ii)\\; for every $(x,y)\\in U$\n\\[\n\\partial_{x}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial_{x}f}{\\partial_{x}g},\\qquad\n\\partial_{y}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial_{y}f}{\\partial_{y}g},\\qquad\n\\partial^{2}_{xy}\\!\\Bigl(\\tfrac{f}{g}\\Bigr)=\n      \\frac{\\partial^{2}_{xy}f}{\\partial^{2}_{xy}g};\n\\]\n\n(iii)\\; $f$ is \\emph{not} a constant multiple of $g$.\n\nEvery admissible couple $(f,g)$ belongs to one and only one of the three\nmutually disjoint classes below.\n\n(A) Separable family.  \nThere exist a base point $(x_{0},y_{0})\\in U$, a constant $C\\neq0$ and\nreal–analytic maps  \n\\[\nr:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0,1\\},\\qquad\ns:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0,1\\},\n\\]\nsuch that  \n\\[\n(s-1)^{2}\\not\\equiv(r-1)^{2},\n\\]\nand  \n\\[\ng(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x} r(t)\\,dt+\n               \\int_{y_{0}}^{y}s(u)\\,du\\Bigr),\\qquad\nf(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x}\\frac{r(t)^{2}}{r(t)-1}\\,dt+\n               \\int_{y_{0}}^{y}\\frac{s(u)^{2}}{s(u)-1}\\,du\\Bigr).\n\\]\n\n(B) Diagonal family.  \nThere exist $C\\neq0$ and a real–analytic map  \n\\[\nh:(0,\\infty)\\longrightarrow\\mathbb R\\setminus\\{0\\},\n\\]\nsuch that for every $t>0$\n\\[\nh'(t)\\neq0,\\quad h''(t)\\neq0,\\quad h'(t)-h(t)\\neq0,\n\\]\nand  \n\\[\ng(x,y)=C\\,h(x+y),\\qquad\nf(x,y)=C\\,h(x+y)\\,\n        \\exp\\!\\Bigl(\\int_{\\,x_{0}+y_{0}}^{\\,x+y}\n                    \\frac{h'(t)}{h'(t)-h(t)}\\,dt\\Bigr),\n\\]\nwhere $(x_{0},y_{0})\\in U$ is fixed once and for all.\n\n(C) Skew–exponential family.  \nThere exist $C\\neq0$, $\\tau_{0}\\in\\mathbb R$ and a real–analytic map  \n\\[\n\\varphi:\\mathbb R\\longrightarrow\\mathbb R\\setminus\\{0\\},\n\\]\nsuch that for every $t\\in\\mathbb R$\n\\[\n\\varphi(t)\\neq0,\\quad\\varphi'(t)\\neq0,\\quad\n\\varphi(t)+\\varphi'(t)\\neq0,\\quad\n2\\varphi(t)+\\varphi'(t)\\neq0,\\quad\n2\\varphi'(t)+\\varphi''(t)\\neq0,\n\\]\nand  \n\\[\ng(x,y)=C\\,\\mathrm e^{2x}\\,\\varphi(x-y),\\qquad\nf(x,y)=C\\,\\mathrm e^{2x}\\,\\varphi(x-y)\\,\n        \\exp\\!\\Bigl(2y+\\int_{\\tau_{0}}^{\\,x-y}\n                     \\frac{2\\varphi(t)+\\varphi'(t)}\n                          {\\varphi(t)+\\varphi'(t)}\\,dt\\Bigr).\n\\]\n\nConversely, every choice of data allowed in {\\rm(A)}, {\\rm(B)} or\n{\\rm(C)} produces an admissible pair, and no other pairs exist.\n\nFinally, exhibit an explicit admissible pair that belongs to {\\rm(C)}\nbut to neither {\\rm(A)} nor {\\rm(B)}.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we write  \n\\[\nu:=\\frac{f}{g},\\qquad\ng_{x}:=\\partial_{x}g,\\;g_{y}:=\\partial_{y}g,\\;\ng_{xy}:=\\partial^{2}_{xy}g,\n\\]\nand use analogous notations for $f$ and $u$.  \nAll differentiations are taken inside $U$.\n\nBecause $f$ and $g$ are real–analytic, every expression constructed from\nthem is real–analytic as well.  This fact will be crucial in\nStep 2 (b).\n\n--------------------------------------------------------------------\n1. Two first–order identities coming from (ii).\n--------------------------------------------------------------------\nRelations (ii) give  \n\\[\nu_{x}=\\frac{f_{x}}{g_{x}},\\qquad\nu_{y}=\\frac{f_{y}}{g_{y}}.\\tag{1}\n\\]\nSince $f=ug$, differentiating yields  \n\\[\nf_{x}=u_{x}g+u\\,g_{x},\\qquad\nf_{y}=u_{y}g+u\\,g_{y}.\\tag{2}\n\\]\nCombining (1)–(2) we arrive at  \n\\[\nu\\,g_{x}=u_{x}\\,(g_{x}-g),\\qquad\nu\\,g_{y}=u_{y}\\,(g_{y}-g).\\tag{3}\n\\]\nBecause $g_{x},g_{y}$ never vanish, we may set  \n\\[\nP:=\\frac{u_{x}}{u}=\\frac{g_{x}}{g_{x}-g},\\qquad\nQ:=\\frac{u_{y}}{u}=\\frac{g_{y}}{g_{y}-g}.\\tag{4}\n\\]\n\n--------------------------------------------------------------------\n2. Two fundamental compatibility relations.\n--------------------------------------------------------------------\n(a) Mixed partials of $u$.  \n\nTaking $\\partial_{y}$ of $P$ and $\\partial_{x}$ of $Q$, using\n$u_{xy}=u_{yx}$ and formulae (3)–(4) one obtains  \n\\[\nP_{y}-Q_{x}= \n\\frac{-g\\,g_{xy}+g_{x}g_{y}}\n     {(g_{x}-g)^{2}(g_{y}-g)^{2}}\n\\Bigl[(g_{y}-g)^{2}-(g_{x}-g)^{2}\\Bigr].\n\\]\nIntroduce  \n\\[\nA:=-g\\,g_{xy}+g_{x}g_{y},\\qquad\nB:=(g_{y}-g)^{2}-(g_{x}-g)^{2}.\\tag{5}\n\\]\nThe equality $u_{xy}=u_{yx}$ is therefore equivalent to  \n\\[\nA\\,B\\equiv0.\\tag{6}\n\\]\n\n(b) The $xy$–condition in (ii).  \n\nThe third relation in (ii) reads\n\\[\nu_{xy}=u\\,\\frac{f_{xy}}{g_{xy}}.\n\\]\nDifferentiate $f=ug$ twice, substitute $f_{xy}=u_{xy}g+u_{x}g_{y}+u_{y}g_{x}+u\\,g_{xy}$,\nand recombine everything with (3).  After a straightforward but lengthy\nsimplification one \\emph{again} obtains the factorisation\n$A\\,B\\equiv0$, identical to (6).\nHence both analytic identities (6) actually coincide.\n\n--------------------------------------------------------------------\n3.  Analytic factorisation implies a dichotomy.\n--------------------------------------------------------------------\nBoth $A$ and $B$ are real–analytic on $U$, hence so is their product.\nBecause $U$ is connected and $A\\,B\\equiv0$, the classical identity\nprinciple yields\n\nEither $A\\equiv0$ on $U$, or $B\\equiv0$ on $U$.   \nIndeed, if $A\\not\\equiv0$, then $A$ is non–vanishing on some open set\n$\\Omega\\subset U$; by (6) we must have $B\\equiv0$ on $\\Omega$, whence\n(analytic continuation) $B\\equiv0$ on all of $U$.  The alternative\n$B\\not\\equiv0$ is treated symmetrically.\n\nThus the dreaded “patchwork’’ possibility is excluded.\n\n--------------------------------------------------------------------\n4. Case $A\\equiv0$ – separable solutions (family {\\rm(A)}).\n--------------------------------------------------------------------\nSet  \n\\[\nr:=\\frac{g_{x}}{g},\\qquad s:=\\frac{g_{y}}{g}.\n\\]\nBecause $A\\equiv0$ one has  \n\\[\nr_{y}=0,\\qquad s_{x}=0,\n\\]\nso $r=r(x)$ and $s=s(y)$.  Solving  \n\\[\n\\frac{g_{x}}{g}=r(x),\\qquad\\frac{g_{y}}{g}=s(y)\n\\]\nalong any path joining $(x_{0},y_{0})$ to $(x,y)$ gives  \n\\[\ng(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x} r(t)\\,dt+\n               \\int_{y_{0}}^{y}s(u)\\,du\\Bigr).\\tag{7}\n\\]\n\nNext, (3)–(4) yield  \n\\[\n\\frac{u_{x}}{u}=\\frac{r}{r-1},\\qquad\n\\frac{u_{y}}{u}=\\frac{s}{s-1},\n\\]\nwhich integrate separately to  \n\\[\nf(x,y)=\nC\\,\\exp\\!\\Bigl(\\int_{x_{0}}^{x}\\frac{r(t)^{2}}{r(t)-1}\\,dt+\n               \\int_{y_{0}}^{y}\\frac{s(u)^{2}}{s(u)-1}\\,du\\Bigr).\\tag{8}\n\\]\n\nFinally  \n\\[\ng_{xy}=r(x)\\,s(y)\\,g(x,y).\\tag{9}\n\\]\nCondition $g_{xy}\\neq0$ together with $g\\neq0$ is therefore equivalent\nto  \n\\[\nr(x)\\neq0\\quad(\\forall x>0),\\qquad\ns(y)\\neq0\\quad(\\forall y>0).\\tag{10}\n\\]\nCollecting (7)–(10), and recalling that $(r-1)^{2}\\not\\equiv\n(s-1)^{2}$ is necessary to avoid $f$ being a scalar multiple of $g$, we\nobtain exactly family {\\rm(A)}.\n\n--------------------------------------------------------------------\n5. Case $A\\not\\equiv0$ – hence $B\\equiv0$.\n--------------------------------------------------------------------\nBecause $g_{x},g_{y}$ never vanish, $B\\equiv0$ rewrites  \n\\[\n\\bigl(g_{y}-g\\bigr)^{2}\\equiv\\bigl(g_{x}-g\\bigr)^{2}.\n\\]\nThus the ratio $(g_{y}-g)/(g_{x}-g)$ is constant, equal to $+1$ or\n$-1$.  Two sub–cases arise.\n\n----------------------------------------------------------\n5.1\\; Sub–case $g_{y}=g_{x}$ – diagonal solutions (family {\\rm(B)}).\n----------------------------------------------------------\nIntroduce $s:=x+y,\\;t:=x-y$.  The equality $g_{y}=g_{x}$ is\n$\\partial_{t}g=0$, so $g(x,y)=h(s)$ for some analytic $h\\not\\equiv0$.\nFormulae (3)–(4) show that $u$ depends on $s$ only.  Integrating in $s$\nyields  \n\\[\nf(x,y)=h(s)\\exp\\!\\Bigl(\\int_{s_{0}}^{s}\n                      \\frac{h'(t)}{h'(t)-h(t)}\\,dt\\Bigr).\\tag{11}\n\\]\nThe non–vanishing requirements in (i) are precisely\n$h',h'',h'-h\\neq0$, giving family {\\rm(B)}.\n\n----------------------------------------------------------\n5.2\\; Sub–case $g_{y}+g_{x}=2g$ – skew–exponential solutions (family {\\rm(C)}).\n----------------------------------------------------------\nSolve the linear first–order PDE $g_{x}+g_{y}=2g$ by characteristics.\nWith $s:=x-y$ one obtains  \n\\[\ng(x,y)=\\Phi(s)\\,\\mathrm e^{2x},\\qquad \\Phi\\in C^{\\omega}(\\mathbb R),\\;\n\\Phi\\neq0.\\tag{12}\n\\]\nEqualities (3)–(4) give  \n\\[\n\\frac{u_{x}}{u}=p(s):=\\frac{2\\Phi+\\Phi'}{\\Phi+\\Phi'},\\qquad\n\\frac{u_{y}}{u}=q(s):=\\frac{\\Phi'}{\\Phi+\\Phi'},\n\\]\nso $p+q\\equiv2$ and $p',q'$ satisfy $p'+q'=0$; the system\n$u_{x}=p(s)u,\\;u_{y}=q(s)u$ is compatible.  Integrating first in $x$,\nthen in $y$ gives  \n\\[\nu(x,y)=D\\,\\mathrm e^{2y}\\,\n        \\exp\\!\\Bigl(\\int_{\\tau_{0}}^{s} p(t)\\,dt\\Bigr).\\tag{13}\n\\]\nPutting $\\varphi:=\\Phi$ and $C:=D$ recovers the formulae of family\n{\\rm(C)}.  The inequalities imposed on $\\varphi$ ensure that the five\nfunctions listed in (i) never vanish.\n\n--------------------------------------------------------------------\n6.  Exhaustiveness and mutual disjointness.\n--------------------------------------------------------------------\nStep 3 provides the global dichotomy $A\\equiv0$ or $B\\equiv0$.  \nThe former gives family {\\rm(A)}; the latter splits into the exclusive\nalternatives $g_{x}=g_{y}$ and $g_{x}+g_{y}=2g$,\nyielding families {\\rm(B)} and {\\rm(C)} respectively.\nConversely, direct substitution shows that each explicit formula in\n{\\rm(A)}, {\\rm(B)} and {\\rm(C)} satisfies conditions (i)–(iii).  \nBecause the three descriptions are pairwise exclusive, the trichotomy is\ncomplete and the classification is unique.\n\n--------------------------------------------------------------------\n7.  An explicit pair lying in {\\rm(C)} only.\n--------------------------------------------------------------------\nChoose  \n\\[\n\\varphi(t):=1+\\mathrm e^{t},\\qquad C:=1,\\qquad\\tau_{0}:=0.\n\\]\nAll required inequalities are obvious:\n$\\varphi>0$, $\\varphi'>0$,  \n$\\varphi+\\varphi' = 1+2\\mathrm e^{t}>0$,  \n$2\\varphi+\\varphi' = 2+3\\mathrm e^{t}>0$,  \n$2\\varphi'+\\varphi'' = 3\\mathrm e^{t}>0$.  \nTherefore  \n\\[\n\\boxed{\\,g(x,y)=\\mathrm e^{2x}\\bigl(1+\\mathrm e^{\\,x-y}\\bigr)}.\n\\]\n\nTo obtain $f$ we compute\n\\[\n\\int_{0}^{\\,x-y}\\frac{2\\varphi(t)+\\varphi'(t)}{\\varphi(t)+\\varphi'(t)}\\,dt\n     =\\int_{0}^{\\,x-y}\\frac{2+3\\mathrm e^{t}}{1+2\\mathrm e^{t}}\\,dt\n     =2(x-y)-\\tfrac12\\log\\!\\bigl(1+2\\mathrm e^{\\,x-y}\\bigr)\n     +\\tfrac12\\log 3.\n\\]\n(The last term is an irrelevant constant that can be absorbed in $C$.)\nHence\n\\[\nu(x,y)=\\mathrm e^{2y}\\,\\exp\\!\\Bigl(2(x-y)-\\tfrac12\\log(1+2\\mathrm e^{\\,x-y})\\Bigr)\n      =\\mathrm e^{2x}\\,(1+2\\mathrm e^{\\,x-y})^{-1/2},\n\\]\nand\n\\[\n\\boxed{\\,\nf(x,y)=\\mathrm e^{2x}\\bigl(1+\\mathrm e^{\\,x-y}\\bigr)\\;\n        \\mathrm e^{2x}\\,(1+2\\mathrm e^{\\,x-y})^{-1/2}}\n      =\\mathrm e^{4x}\\,\n        \\frac{1+\\mathrm e^{\\,x-y}}{\\sqrt{1+2\\mathrm e^{\\,x-y}}}\\;.\n\\]\n\nSince $\\mathrm e^{\\,x-y}>0$ on $U$, the square root is real–analytic,\nand one checks directly that $g_{x},g_{y},g_{xy}$ never vanish.\nMoreover the pair is obviously neither separable nor diagonal; hence it\nbelongs to {\\rm(C)} and to neither {\\rm(A)} nor {\\rm(B)}.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.380814",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension: the problem moves from one variable to two variables, forcing the use of partial derivatives and mixed partials.\n2.  Additional constraints: three simultaneous identities (two first-order and one second-order) must hold, greatly restricting admissible functions.\n3.  Sophisticated structure: solving requires analysing an over-determined nonlinear PDE system, reducing it to an algebraic system, and checking global non-vanishing conditions.\n4.  Deeper theory: the exhaustiveness proof invokes separation of variables, the structure of solutions of the original ODE, and compatibility of cross-derivatives.\n5.  Multiple interacting concepts: one must synchronise conditions in x and y directions and for mixed derivatives; any naive “copy” of the 1-D example fails unless the parameters are tuned to satisfy all three algebraic equations.\n\nHence the variant is substantially harder than both the original problem and the current kernel variant, demanding multi-variable calculus, PDE reasoning, and a careful algebraic compatibility analysis."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}