{
  "index": "1983-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem B-3\n\nAssume that the differential equation\n\\[\ny^{\\prime \\prime \\prime}+p(x) y^{\\prime \\prime}+q(x) y^{\\prime}+r(x) y=0\n\\]\nhas solutions \\( y_{1}(x), y_{2}(x) \\), and \\( y_{3}(x) \\) on the whole real line such that\n\\[\ny_{1}^{2}(x)+y_{2}^{2}(x)+y_{3}^{2}(x)=1\n\\]\nfor all real \\( x \\). Let\n\\[\nf(x)=\\left(y_{1}^{\\prime}(x)\\right)^{2}+\\left(y_{2}^{\\prime}(x)\\right)^{2}+\\left(y_{3}^{\\prime}(x)\\right)^{2} .\n\\]\n\nFind constants \\( A \\) and \\( B \\) such that \\( f(x) \\) is a solution to the differential equation\n\\[\ny^{\\prime}+A p(x) y=\\operatorname{Br}(x)\n\\]",
  "solution": "B-3.\nTo satisfy the equation, each \\( y_{i} \\) must have at least 3 derivatives. Here \\( \\Sigma \\) will be a sum with \\( i \\) running over \\( 1,2,3 \\). We have \\( \\Sigma y_{i}^{2}=1 \\) and \\( \\Sigma\\left(y_{i}^{\\prime}\\right)^{2}=f \\). Differentiating, one has \\( \\Sigma 2 y_{i} y_{i}^{\\prime}=0 \\) and \\( \\Sigma 2 y_{i}^{\\prime} y_{i}^{\\prime \\prime}=f^{\\prime} \\). Differentiating \\( \\Sigma y_{i} y_{i}^{\\prime}=0 \\) leads to \\( \\Sigma y_{i} y_{i}^{\\prime \\prime}+\\Sigma\\left(y_{i}^{\\prime}\\right)^{2}=0 \\) so \\( \\Sigma y_{i} y_{i}^{\\prime \\prime}=-f \\). Differentiating this gives us \\( \\Sigma y_{i}^{\\prime} y_{i}^{\\prime \\prime}+\\sum y_{i} y_{i}^{\\prime \\prime \\prime}=-f^{\\prime} \\). This and \\( \\Sigma y_{i}^{\\prime} y_{i}^{\\prime \\prime}=f^{\\prime} / 2 \\) leads to \\( \\Sigma y_{i} y_{i}^{\\prime \\prime \\prime}=-3 f^{\\prime} / 2 \\). Multiplying each term of\n\\[\ny_{i}^{\\prime \\prime \\prime}+p y_{i}^{\\prime \\prime}+q y_{i}^{\\prime}+r y_{i}=0\n\\]\nby \\( y_{i} \\) and summing gives us\n\\[\n-3 f^{\\prime} / 2-p f+q \\cdot 0+r=0\n\\]",
  "vars": [
    "x",
    "y",
    "y_i",
    "y_1",
    "y_2",
    "y_3",
    "f"
  ],
  "params": [
    "p",
    "q",
    "r",
    "A",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "y": "function",
        "y_i": "indexfn",
        "y_1": "firstfn",
        "y_2": "secondf",
        "y_3": "thirdfn",
        "f": "derivsum",
        "p": "coeffic",
        "q": "coefsec",
        "r": "coefthe",
        "A": "constan",
        "B": "paramtr"
      },
      "question": "Problem B-3\n\nAssume that the differential equation\n\\[\nfunction^{\\prime \\prime \\prime}+coeffic(variable)\\ function^{\\prime \\prime}+coefsec(variable)\\ function^{\\prime}+coefthe(variable)\\ function=0\n\\]\nhas solutions \\( firstfn(variable), secondf(variable) \\), and \\( thirdfn(variable) \\) on the whole real line such that\n\\[\nfirstfn^{2}(variable)+secondf^{2}(variable)+thirdfn^{2}(variable)=1\n\\]\nfor all real \\( variable \\). Let\n\\[\nderivsum(variable)=\\left(firstfn^{\\prime}(variable)\\right)^{2}+\\left(secondf^{\\prime}(variable)\\right)^{2}+\\left(thirdfn^{\\prime}(variable)\\right)^{2} .\n\\]\n\nFind constants \\( constan \\) and \\( paramtr \\) such that \\( derivsum(variable) \\) is a solution to the differential equation\n\\[\nfunction^{\\prime}+constan\\ coeffic(variable)\\ function=paramtr\\ coefthe(variable)\n\\]",
      "solution": "B-3.\nTo satisfy the equation, each \\( indexfn \\) must have at least 3 derivatives. Here \\( \\Sigma \\) will be a sum with \\( i \\) running over \\( 1,2,3 \\). We have \\( \\Sigma indexfn^{2}=1 \\) and \\( \\Sigma\\left(indexfn^{\\prime}\\right)^{2}=derivsum \\). Differentiating, one has \\( \\Sigma 2\\ indexfn\\ indexfn^{\\prime}=0 \\) and \\( \\Sigma 2\\ indexfn^{\\prime}\\ indexfn^{\\prime \\prime}=derivsum^{\\prime} \\). Differentiating \\( \\Sigma indexfn\\ indexfn^{\\prime}=0 \\) leads to \\( \\Sigma indexfn\\ indexfn^{\\prime \\prime}+\\Sigma\\left(indexfn^{\\prime}\\right)^{2}=0 \\) so \\( \\Sigma indexfn\\ indexfn^{\\prime \\prime}=-derivsum \\). Differentiating this gives us \\( \\Sigma indexfn^{\\prime}\\ indexfn^{\\prime \\prime}+\\sum indexfn\\ indexfn^{\\prime \\prime \\prime}=-derivsum^{\\prime} \\). This and \\( \\Sigma indexfn^{\\prime}\\ indexfn^{\\prime \\prime}=derivsum^{\\prime} / 2 \\) leads to \\( \\Sigma indexfn\\ indexfn^{\\prime \\prime \\prime}=-3\\ derivsum^{\\prime} / 2 \\). Multiplying each term of\n\\[\nindexfn^{\\prime \\prime \\prime}+coeffic\\ indexfn^{\\prime \\prime}+coefsec\\ indexfn^{\\prime}+coefthe\\ indexfn=0\n\\]\nby \\( indexfn \\) and summing gives us\n\\[\n-3\\ derivsum^{\\prime} / 2-coeffic\\ derivsum+coefsec \\cdot 0+coefthe=0\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "teaspoon",
        "y_i": "cheesecake",
        "y_1": "marigold",
        "y_2": "snowflake",
        "y_3": "pinecone",
        "f": "drumstick",
        "p": "whirlwind",
        "q": "buttercup",
        "r": "arrowhead",
        "A": "sandpaper",
        "B": "treetruck"
      },
      "question": "Problem B-3\n\nAssume that the differential equation\n\\[\nteaspoon^{\\prime \\prime \\prime}+whirlwind(lighthouse) teaspoon^{\\prime \\prime}+buttercup(lighthouse) teaspoon^{\\prime}+arrowhead(lighthouse) teaspoon=0\n\\]\nhas solutions \\( marigold(lighthouse), snowflake(lighthouse) \\), and \\( pinecone(lighthouse) \\) on the whole real line such that\n\\[\nmarigold^{2}(lighthouse)+snowflake^{2}(lighthouse)+pinecone^{2}(lighthouse)=1\n\\]\nfor all real \\( lighthouse \\). Let\n\\[\ndrumstick(lighthouse)=\\left(marigold^{\\prime}(lighthouse)\\right)^{2}+\\left(snowflake^{\\prime}(lighthouse)\\right)^{2}+\\left(pinecone^{\\prime}(lighthouse)\\right)^{2} .\n\\]\n\nFind constants \\( sandpaper \\) and \\( treetruck \\) such that \\( drumstick(lighthouse) \\) is a solution to the differential equation\n\\[\nteaspoon^{\\prime}+sandpaper\\, whirlwind(lighthouse)\\, teaspoon=\\operatorname{Br}(lighthouse)\n\\]",
      "solution": "B-3.\nTo satisfy the equation, each \\( cheesecake \\) must have at least 3 derivatives. Here \\( \\Sigma \\) will be a sum with \\( i \\) running over \\( 1,2,3 \\). We have \\( \\Sigma cheesecake^{2}=1 \\) and \\( \\Sigma\\left(cheesecake^{\\prime}\\right)^{2}=drumstick \\). Differentiating, one has \\( \\Sigma 2 cheesecake\\, cheesecake^{\\prime}=0 \\) and \\( \\Sigma 2 cheesecake^{\\prime}\\, cheesecake^{\\prime \\prime}=drumstick^{\\prime} \\). Differentiating \\( \\Sigma cheesecake\\, cheesecake^{\\prime}=0 \\) leads to \\( \\Sigma cheesecake\\, cheesecake^{\\prime \\prime}+\\Sigma\\left(cheesecake^{\\prime}\\right)^{2}=0 \\) so \\( \\Sigma cheesecake\\, cheesecake^{\\prime \\prime}=-drumstick \\). Differentiating this gives us \\( \\Sigma cheesecake^{\\prime}\\, cheesecake^{\\prime \\prime}+\\sum cheesecake\\, cheesecake^{\\prime \\prime \\prime}=-drumstick^{\\prime} \\). This and \\( \\Sigma cheesecake^{\\prime}\\, cheesecake^{\\prime \\prime}=drumstick^{\\prime} / 2 \\) leads to \\( \\Sigma cheesecake\\, cheesecake^{\\prime \\prime \\prime}=-3\\, drumstick^{\\prime} / 2 \\). Multiplying each term of\n\\[\ncheesecake^{\\prime \\prime \\prime}+whirlwind\\, cheesecake^{\\prime \\prime}+buttercup\\, cheesecake^{\\prime}+arrowhead\\, cheesecake=0\n\\]\nby \\( cheesecake \\) and summing gives us\n\\[\n-3\\, drumstick^{\\prime} / 2-\\, whirlwind\\, drumstick+buttercup \\cdot 0+arrowhead=0\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stillpoint",
        "y": "nonoutput",
        "y_i": "nonindex",
        "y_1": "lastsoln",
        "y_2": "middlesol",
        "y_3": "initialsol",
        "f": "constancy",
        "p": "staticfun",
        "q": "tranquilx",
        "r": "calmness",
        "A": "variable",
        "B": "shifting"
      },
      "question": "Problem B-3\n\nAssume that the differential equation\n\\[\nnonoutput^{\\prime \\prime \\prime}+staticfun(stillpoint)\\,nonoutput^{\\prime \\prime}+tranquilx(stillpoint)\\,nonoutput^{\\prime}+calmness(stillpoint)\\,nonoutput=0\n\\]\nhas solutions \\( lastsoln(stillpoint), middlesol(stillpoint), \\) and \\( initialsol(stillpoint) \\) on the whole real line such that\n\\[\nlastsoln^{2}(stillpoint)+middlesol^{2}(stillpoint)+initialsol^{2}(stillpoint)=1\n\\]\nfor all real \\( stillpoint \\). Let\n\\[\nconstancy(stillpoint)=\\left(lastsoln^{\\prime}(stillpoint)\\right)^{2}+\\left(middlesol^{\\prime}(stillpoint)\\right)^{2}+\\left(initialsol^{\\prime}(stillpoint)\\right)^{2} .\n\\]\n\nFind constants \\( variable \\) and \\( shifting \\) such that \\( constancy(stillpoint) \\) is a solution to the differential equation\n\\[\nnonoutput^{\\prime}+variable\\,staticfun(stillpoint)\\,nonoutput=\\operatorname{Br}(stillpoint)\n\\]",
      "solution": "B-3.\nTo satisfy the equation, each \\( nonindex \\) must have at least 3 derivatives. Here \\( \\Sigma \\) will be a sum with \\( i \\) running over \\( 1,2,3 \\). We have \\( \\Sigma nonindex^{2}=1 \\) and \\( \\Sigma\\left(nonindex^{\\prime}\\right)^{2}=constancy \\). Differentiating, one has \\( \\Sigma 2\\,nonindex\\,nonindex^{\\prime}=0 \\) and \\( \\Sigma 2\\,nonindex^{\\prime}\\,nonindex^{\\prime \\prime}=constancy^{\\prime} \\). Differentiating \\( \\Sigma nonindex\\,nonindex^{\\prime}=0 \\) leads to \\( \\Sigma nonindex\\,nonindex^{\\prime \\prime}+\\Sigma\\left(nonindex^{\\prime}\\right)^{2}=0 \\) so \\( \\Sigma nonindex\\,nonindex^{\\prime \\prime}=-constancy \\). Differentiating this gives us \\( \\Sigma nonindex^{\\prime}\\,nonindex^{\\prime \\prime}+\\sum nonindex\\,nonindex^{\\prime \\prime \\prime}=-constancy^{\\prime} \\). This and \\( \\Sigma nonindex^{\\prime}\\,nonindex^{\\prime \\prime}=constancy^{\\prime} / 2 \\) leads to \\( \\Sigma nonindex\\,nonindex^{\\prime \\prime \\prime}=-3\\,constancy^{\\prime} / 2 \\). Multiplying each term of\n\\[\nnonindex^{\\prime \\prime \\prime}+staticfun\\,nonindex^{\\prime \\prime}+tranquilx\\,nonindex^{\\prime}+calmness\\,nonindex=0\n\\]\nby \\( nonindex \\) and summing gives us\n\\[\n-3\\,constancy^{\\prime} / 2-staticfun\\,constancy+tranquilx \\cdot 0+calmness=0\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "yzsdmnlk",
        "y": "qrefplaz",
        "y_i": "bajsklwe",
        "y_1": "hvjdskea",
        "y_2": "plamkqrz",
        "y_3": "xzvwhlpo",
        "f": "cnavpoiu",
        "p": "rtyusdfg",
        "q": "wernxspo",
        "r": "lkfdjsam",
        "A": "pvnjkliu",
        "B": "zcxvqmwe"
      },
      "question": "Problem B-3\n\nAssume that the differential equation\n\\[\nqrefplaz^{\\prime \\prime \\prime}+rtyusdfg(yzsdmnlk) qrefplaz^{\\prime \\prime}+wernxspo(yzsdmnlk) qrefplaz^{\\prime}+lkfdjsam(yzsdmnlk) qrefplaz=0\n\\]\nhas solutions \\( hvjdskea(yzsdmnlk), plamkqrz(yzsdmnlk), \\) and \\( xzvwhlpo(yzsdmnlk) \\) on the whole real line such that\n\\[\nhvjdskea^{2}(yzsdmnlk)+plamkqrz^{2}(yzsdmnlk)+xzvwhlpo^{2}(yzsdmnlk)=1\n\\]\nfor all real \\( yzsdmnlk \\). Let\n\\[\ncnavpoiu(yzsdmnlk)=\\left(hvjdskea^{\\prime}(yzsdmnlk)\\right)^{2}+\\left(plamkqrz^{\\prime}(yzsdmnlk)\\right)^{2}+\\left(xzvwhlpo^{\\prime}(yzsdmnlk)\\right)^{2} .\n\\]\n\nFind constants \\( pvnjkliu \\) and \\( zcxvqmwe \\) such that \\( cnavpoiu(yzsdmnlk) \\) is a solution to the differential equation\n\\[\nqrefplaz^{\\prime}+pvnjkliu rtyusdfg(yzsdmnlk) qrefplaz=\\operatorname{zcxvqmwelkfdjsam}(yzsdmnlk)\n\\]",
      "solution": "B-3.\nTo satisfy the equation, each \\( bajsklwe_{i} \\) must have at least 3 derivatives. Here \\( \\Sigma \\) will be a sum with \\( i \\) running over \\( 1,2,3 \\). We have \\( \\Sigma bajsklwe_{i}^{2}=1 \\) and \\( \\Sigma\\left(bajsklwe_{i}^{\\prime}\\right)^{2}=cnavpoiu \\). Differentiating, one has \\( \\Sigma 2 bajsklwe_{i} bajsklwe_{i}^{\\prime}=0 \\) and \\( \\Sigma 2 bajsklwe_{i}^{\\prime} bajsklwe_{i}^{\\prime \\prime}=cnavpoiu^{\\prime} \\). Differentiating \\( \\Sigma bajsklwe_{i} bajsklwe_{i}^{\\prime}=0 \\) leads to \\( \\Sigma bajsklwe_{i} bajsklwe_{i}^{\\prime \\prime}+\\Sigma\\left(bajsklwe_{i}^{\\prime}\\right)^{2}=0 \\) so \\( \\Sigma bajsklwe_{i} bajsklwe_{i}^{\\prime \\prime}=-cnavpoiu \\). Differentiating this gives us \\( \\Sigma bajsklwe_{i}^{\\prime} bajsklwe_{i}^{\\prime \\prime}+\\sum bajsklwe_{i} bajsklwe_{i}^{\\prime \\prime \\prime}=-cnavpoiu^{\\prime} \\). This and \\( \\Sigma bajsklwe_{i}^{\\prime} bajsklwe_{i}^{\\prime \\prime}=cnavpoiu^{\\prime} / 2 \\) leads to \\( \\Sigma bajsklwe_{i} bajsklwe_{i}^{\\prime \\prime \\prime}=-3 cnavpoiu^{\\prime} / 2 \\). Multiplying each term of\n\\[\nbajsklwe_{i}^{\\prime \\prime \\prime}+rtyusdfg bajsklwe_{i}^{\\prime \\prime}+wernxspo bajsklwe_{i}^{\\prime}+lkfdjsam bajsklwe_{i}=0\n\\]\nby \\( bajsklwe_{i} \\) and summing gives us\n\\[\n-3 cnavpoiu^{\\prime} / 2-rtyusdfg cnavpoiu+wernxspo \\cdot 0+lkfdjsam=0\n\\]\n"
    },
    "kernel_variant": {
      "question": "Consider the fourth–order linear differential equation    \n    y^{(4)}(x)+p(x)\\,y^{(3)}(x)+q(x)\\,y''(x)+r(x)\\,y'(x)+s(x)\\,y(x)=0   (1)\n\non the whole real line.  \nThroughout we assume  \n\n• p,q,r,s∈C¹(ℝ),  \n• (1) possesses four linearly independent solutions y₁,y₂,y₃,y₄∈C⁴(ℝ),  \n\nand that these solutions satisfy the prescribed “radius–four’’ constraint  \n    y₁(x)²+y₂(x)²+y₃(x)²+y₄(x)² = 16   for every x∈ℝ.    (2)\n\nIntroduce the quadratic energy functions  \n    f(x)=Σ_{i=1}^{4} (y_i'(x))²,  g(x)=Σ_{i=1}^{4} (y_i''(x))².   (3)\n\n(a) Show that there exist universal numerical constants A,B,C,D (i.e. independent of the coefficient functions p,q,r,s) such that  \n    g(x)=A\\,f''(x)+B\\,p(x)\\,f'(x)+C\\,q(x)\\,f(x)+D\\,s(x),   (4)  \nand determine A,B,C,D explicitly.\n\n(b) Differentiate (4) once and, using a second identity that follows from (1), eliminate g and g′ to obtain a third–order inhomogeneous linear differential equation for f alone,  \n    f'''(x)+α₁(x)f''(x)+α₂(x)f'(x)+α₃(x)f(x)=β(x),   (5)  \nwhere  \n\n    α₁(x)= 3 p(x)/2,  \n\n    α₂(x)= 9 p′(x)/10 + 3 p(x)²/5 + 2 q(x)/5,  \n\n    α₃(x)= 3 q′(x)/5 + 2 p(x)q(x)/5 − 2 r(x)/5,  \n\n    β(x)= (16/5) [ 3 s′(x) + 2 p(x)s(x) ] .   (6)\n\nGive a complete, carefully justified proof of (5).",
      "solution": "Throughout Σ denotes Σ_{i=1}^{4}.  Because the four solutions are C⁴ and the coefficients are C¹, every derivative that occurs below exists and is continuous on ℝ.\n\n--------------------------------------------------------------------\n1.  Algebraic identities forced by the constraint\n--------------------------------------------------------------------\nSet  \n\n    Y(x)=(y₁(x),y₂(x),y₃(x),y₄(x))∈ℝ⁴,  ‖Y‖²=16.\n\nDefine, besides f and g in (3),\n\n    U(x)=Σ y_i′y_i″,  X(x)=Σ y_i′y_i‴,  Z(x)=Σ y_i″y_i‴.  (7)\n\n(a) First derivative of the constraint  \n    d/dx‖Y‖² = 2 Σ y_i y_i′ = 0 ⇒ Σ y_i y_i′ = 0.   (8)\n\n(b) Second derivative of the constraint  \n    d/dx(Σ y_i y_i′) = Σ (y_i′² + y_i y_i″) = f + Σ y_i y_i″ = 0  \n    ⇒ Σ y_i y_i″ = −f.              (9)\n\n(c) Third derivative of the constraint  \nDifferentiate (9):\n\n    Σ (y_i′ y_i″ + y_i y_i‴) = −f′.       (10)\n\nBut Σ y_i′ y_i″ = U, so (10) reads U + Σ y_i y_i‴ = −f′.  \nSince f′ = 2U, we obtain\n\n    Σ y_i y_i‴ = −(3/2) f′.         (11)\n\n(d) Relations between f, g, U and X  \nFrom the definition of f,\n\n    f′ = 2 Σ y_i′ y_i″ = 2U ⇒ U = f′/2.   (12)\n\nDifferentiating U gives\n\n    U′ = Σ (y_i″² + y_i′ y_i‴) = g + X.   (13)\n\nHence\n\n    X = U′ − g = f″/2 − g.        (14)\n\n--------------------------------------------------------------------\n2.  Proof of the identity (4)\n--------------------------------------------------------------------\nMultiply (1) by y_i, sum over i and use (8)–(11):\n\n    Σ y_i y_i^{(4)} + p Σ y_i y_i‴ + q Σ y_i y_i″ + r Σ y_i y_i′ + s Σ y_i² = 0\n    ⇒ Σ y_i y_i^{(4)} − (3/2) p f′ − q f + 16 s = 0.   (15)\n\nTo eliminate Σ y_i y_i^{(4)} apply (14).  \nDifferentiate (11):\n\n    (Σ y_i y_i‴)' = Σ y_i′ y_i‴ + Σ y_i y_i^{(4)} = −(3/2)f″,  \n\nso that  \n\n    Σ y_i y_i^{(4)} = −(3/2) f″ − X = −2 f″ + g  (by (14)). (16)\n\nInsert (16) into (15):\n\n    g = 2 f″ + (3/2) p f′ + q f − 16 s.      (17)\n\nThus\n\n    A = 2, B = 3/2, C = 1, D = −16,\n\nwhich proves part (a).\n\n--------------------------------------------------------------------\n3.  A second identity linking g′ to f and its derivatives\n--------------------------------------------------------------------\nMultiply (1) by y_i′ and sum:\n\n    Σ y_i′ y_i^{(4)} + p Σ y_i′ y_i‴ + q Σ y_i′ y_i″ + r Σ y_i′² + s Σ y_i y_i′ = 0.\n\nUsing (8) the last term vanishes; in the notation (7),\n\n    Σ y_i′ y_i^{(4)} + p X + q U + r f = 0.    (18)\n\nDifferentiate X (cf. (7)):\n\n    X′ = Σ (y_i″ y_i‴ + y_i′ y_i^{(4)}) = Z + Σ y_i′ y_i^{(4)}.  (19)\n\nBecause g′ = 2Z, equation (18) together with (19) yields\n\n    X′ − g′/2 + p X + (q/2) f′ + r f = 0.   (20)\n\nNow substitute X = f″/2 − g from (14) and the expressions\n\n    g  = 2 f″ + (3/2) p f′ + q f − 16 s,  \n    g′ = 2 f‴ + (3/2)(p′ f′ + p f″) + q′ f + q f′ − 16 s′.\n\nAfter a straightforward but careful simplification one finds\n\n    −(5/2) f‴ − (15/4) p f″ − (9/4) p′ f′ − (3/2) p² f′ − q f′  \n      − (3/2) q′ f − p q f + r f + 24 s′ + 16 p s = 0.  (21)\n\n--------------------------------------------------------------------\n4.  The differential equation for f\n--------------------------------------------------------------------\nMultiply (21) by −4 and divide by 10 to normalise the leading coefficient to +1; the terms containing s and s′ move to the right–hand side:\n\n    f‴ + (3/2) p f″  \n     + (9 p′/10 + 3 p²/5 + 2 q/5) f′  \n     + (3 q′/5 + 2 p q/5 − 2 r/5) f  \n    = (48/5) s′ + (32/5) p s.   (22)\n\nEquation (22) is exactly (5)–(6) with\n\n    α₁ = 3p/2,  \n    α₂ = 9p′/10 + 3p²/5 + 2q/5,  \n    α₃ = 3q′/5 + 2pq/5 − 2r/5,  \n    β  = (16/5)(3s′ + 2ps).\n\nHence both parts (a) and (b) are established.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.671899",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher order & more variables: the equation is raised from 3rd to 4th order and the number of interacting solutions from three to four, doubling the ambient dimension of the geometric constraint.  \n2. Additional dependent quantities: besides f(x) the problem introduces g(x)=Σy_i''² and forces the solver to manage their interplay.  \n3. Deeper algebraic structure: the derivation requires a chain of nested scalar identities among higher–order mixed products of the solutions (U,V,X,Z) and repeated differentiation of these identities.  \n4. Elimination technique: one must skillfully eliminate g by combining its explicit expression with a first-order differential relation, leading to a third-order inhomogeneous ODE for f — far subtler than the first-order relation in the original problem.  \n5. Universality: the constants A,B,C,D are shown to be independent of the coefficient functions, demanding a careful bookkeeping of every term.  \n\nAll of these elements significantly raise the technical level beyond the original problem, necessitating multiple advanced techniques including vector–valued calculus, systematic differentiation of orthogonality relations, and elimination of auxiliary variables."
      }
    },
    "original_kernel_variant": {
      "question": "Consider the fourth–order linear differential equation    \n    y^{(4)}(x)+p(x)\\,y^{(3)}(x)+q(x)\\,y''(x)+r(x)\\,y'(x)+s(x)\\,y(x)=0   (1)\n\non the whole real line.  \nThroughout we assume  \n\n• p,q,r,s∈C¹(ℝ),  \n• (1) possesses four linearly independent solutions y₁,y₂,y₃,y₄∈C⁴(ℝ),  \n\nand that these solutions satisfy the prescribed “radius–four’’ constraint  \n    y₁(x)²+y₂(x)²+y₃(x)²+y₄(x)² = 16   for every x∈ℝ.    (2)\n\nIntroduce the quadratic energy functions  \n    f(x)=Σ_{i=1}^{4} (y_i'(x))²,  g(x)=Σ_{i=1}^{4} (y_i''(x))².   (3)\n\n(a) Show that there exist universal numerical constants A,B,C,D (i.e. independent of the coefficient functions p,q,r,s) such that  \n    g(x)=A\\,f''(x)+B\\,p(x)\\,f'(x)+C\\,q(x)\\,f(x)+D\\,s(x),   (4)  \nand determine A,B,C,D explicitly.\n\n(b) Differentiate (4) once and, using a second identity that follows from (1), eliminate g and g′ to obtain a third–order inhomogeneous linear differential equation for f alone,  \n    f'''(x)+α₁(x)f''(x)+α₂(x)f'(x)+α₃(x)f(x)=β(x),   (5)  \nwhere  \n\n    α₁(x)= 3 p(x)/2,  \n\n    α₂(x)= 9 p′(x)/10 + 3 p(x)²/5 + 2 q(x)/5,  \n\n    α₃(x)= 3 q′(x)/5 + 2 p(x)q(x)/5 − 2 r(x)/5,  \n\n    β(x)= (16/5) [ 3 s′(x) + 2 p(x)s(x) ] .   (6)\n\nGive a complete, carefully justified proof of (5).",
      "solution": "Throughout Σ denotes Σ_{i=1}^{4}.  Because the four solutions are C⁴ and the coefficients are C¹, every derivative that occurs below exists and is continuous on ℝ.\n\n--------------------------------------------------------------------\n1.  Algebraic identities forced by the constraint\n--------------------------------------------------------------------\nSet  \n\n    Y(x)=(y₁(x),y₂(x),y₃(x),y₄(x))∈ℝ⁴,  ‖Y‖²=16.\n\nDefine, besides f and g in (3),\n\n    U(x)=Σ y_i′y_i″,  X(x)=Σ y_i′y_i‴,  Z(x)=Σ y_i″y_i‴.  (7)\n\n(a) First derivative of the constraint  \n    d/dx‖Y‖² = 2 Σ y_i y_i′ = 0 ⇒ Σ y_i y_i′ = 0.   (8)\n\n(b) Second derivative of the constraint  \n    d/dx(Σ y_i y_i′) = Σ (y_i′² + y_i y_i″) = f + Σ y_i y_i″ = 0  \n    ⇒ Σ y_i y_i″ = −f.              (9)\n\n(c) Third derivative of the constraint  \nDifferentiate (9):\n\n    Σ (y_i′ y_i″ + y_i y_i‴) = −f′.       (10)\n\nBut Σ y_i′ y_i″ = U, so (10) reads U + Σ y_i y_i‴ = −f′.  \nSince f′ = 2U, we obtain\n\n    Σ y_i y_i‴ = −(3/2) f′.         (11)\n\n(d) Relations between f, g, U and X  \nFrom the definition of f,\n\n    f′ = 2 Σ y_i′ y_i″ = 2U ⇒ U = f′/2.   (12)\n\nDifferentiating U gives\n\n    U′ = Σ (y_i″² + y_i′ y_i‴) = g + X.   (13)\n\nHence\n\n    X = U′ − g = f″/2 − g.        (14)\n\n--------------------------------------------------------------------\n2.  Proof of the identity (4)\n--------------------------------------------------------------------\nMultiply (1) by y_i, sum over i and use (8)–(11):\n\n    Σ y_i y_i^{(4)} + p Σ y_i y_i‴ + q Σ y_i y_i″ + r Σ y_i y_i′ + s Σ y_i² = 0\n    ⇒ Σ y_i y_i^{(4)} − (3/2) p f′ − q f + 16 s = 0.   (15)\n\nTo eliminate Σ y_i y_i^{(4)} apply (14).  \nDifferentiate (11):\n\n    (Σ y_i y_i‴)' = Σ y_i′ y_i‴ + Σ y_i y_i^{(4)} = −(3/2)f″,  \n\nso that  \n\n    Σ y_i y_i^{(4)} = −(3/2) f″ − X = −2 f″ + g  (by (14)). (16)\n\nInsert (16) into (15):\n\n    g = 2 f″ + (3/2) p f′ + q f − 16 s.      (17)\n\nThus\n\n    A = 2, B = 3/2, C = 1, D = −16,\n\nwhich proves part (a).\n\n--------------------------------------------------------------------\n3.  A second identity linking g′ to f and its derivatives\n--------------------------------------------------------------------\nMultiply (1) by y_i′ and sum:\n\n    Σ y_i′ y_i^{(4)} + p Σ y_i′ y_i‴ + q Σ y_i′ y_i″ + r Σ y_i′² + s Σ y_i y_i′ = 0.\n\nUsing (8) the last term vanishes; in the notation (7),\n\n    Σ y_i′ y_i^{(4)} + p X + q U + r f = 0.    (18)\n\nDifferentiate X (cf. (7)):\n\n    X′ = Σ (y_i″ y_i‴ + y_i′ y_i^{(4)}) = Z + Σ y_i′ y_i^{(4)}.  (19)\n\nBecause g′ = 2Z, equation (18) together with (19) yields\n\n    X′ − g′/2 + p X + (q/2) f′ + r f = 0.   (20)\n\nNow substitute X = f″/2 − g from (14) and the expressions\n\n    g  = 2 f″ + (3/2) p f′ + q f − 16 s,  \n    g′ = 2 f‴ + (3/2)(p′ f′ + p f″) + q′ f + q f′ − 16 s′.\n\nAfter a straightforward but careful simplification one finds\n\n    −(5/2) f‴ − (15/4) p f″ − (9/4) p′ f′ − (3/2) p² f′ − q f′  \n      − (3/2) q′ f − p q f + r f + 24 s′ + 16 p s = 0.  (21)\n\n--------------------------------------------------------------------\n4.  The differential equation for f\n--------------------------------------------------------------------\nMultiply (21) by −4 and divide by 10 to normalise the leading coefficient to +1; the terms containing s and s′ move to the right–hand side:\n\n    f‴ + (3/2) p f″  \n     + (9 p′/10 + 3 p²/5 + 2 q/5) f′  \n     + (3 q′/5 + 2 p q/5 − 2 r/5) f  \n    = (48/5) s′ + (32/5) p s.   (22)\n\nEquation (22) is exactly (5)–(6) with\n\n    α₁ = 3p/2,  \n    α₂ = 9p′/10 + 3p²/5 + 2q/5,  \n    α₃ = 3q′/5 + 2pq/5 − 2r/5,  \n    β  = (16/5)(3s′ + 2ps).\n\nHence both parts (a) and (b) are established.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.527028",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher order & more variables: the equation is raised from 3rd to 4th order and the number of interacting solutions from three to four, doubling the ambient dimension of the geometric constraint.  \n2. Additional dependent quantities: besides f(x) the problem introduces g(x)=Σy_i''² and forces the solver to manage their interplay.  \n3. Deeper algebraic structure: the derivation requires a chain of nested scalar identities among higher–order mixed products of the solutions (U,V,X,Z) and repeated differentiation of these identities.  \n4. Elimination technique: one must skillfully eliminate g by combining its explicit expression with a first-order differential relation, leading to a third-order inhomogeneous ODE for f — far subtler than the first-order relation in the original problem.  \n5. Universality: the constants A,B,C,D are shown to be independent of the coefficient functions, demanding a careful bookkeeping of every term.  \n\nAll of these elements significantly raise the technical level beyond the original problem, necessitating multiple advanced techniques including vector–valued calculus, systematic differentiation of orthogonality relations, and elimination of auxiliary variables."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}